2010 June « The Unapologetic Mathematician

Fatou’s Lemma

Today we prove Fatou’s Lemma, which is a precursor to the Fatou-Lebesgue theorem, and an important result in its own right.

If $\{f_n\}$ is a sequence of non-negative integrable functions then the function defined pointwise as

$\displaystyle f_*(x)=\liminf\limits_{n\to\infty} f_n(x)$

is also integrable, and we have the inequality

$\displaystyle\int f_*\,d\mu\leq\liminf\limits_{n\to\infty}\int f_n\,d\mu$

In fact, the lemma is often stated for a sequence of measurable functions and concludes that $f_*$ is measurable (along with the inequality), but we already know that the limit inferior of a sequence of measurable functions is measurable, and so the integrable case is the most interesting part for us.

So, we define the functions

$\displaystyle g_n(x)=\inf\limits_{i\geq n}f_i(x)$

so that each $g_n$ is integrable, each $g_n\leq f_n$ and the sequence $\{g_n\}$ is pointwise increasing. Monotonicity tells us that for each $n$ we have

$\displaystyle\int g_n\,d\mu\leq\int f_n\,d\mu$

and it follows that

$\displaystyle\lim\limits_{n\to\infty}\int g_n\,d\mu\leq\liminf\limits_{n\to\infty}\int f_n\,d\mu<\infty$

We also know that

$\displaystyle\lim\limits_{n\to\infty}g_n(x)=\liminf\limits_{n\to\infty}f_n(x)=f_*(x)$

which means we can bring the monotone convergence theorem to bear. This tells us that

$\displaystyle\int f_*\,d\mu=\lim\limits_{n\to\infty}\int g_n\,d\mu\leq\liminf\limits_{n\to\infty}\int f_n\,d\mu$

as asserted.

If it happened that $f_*$ were not integrable, then some of the $f_n$ would have to be only measurable — not integrable — themselves. And it couldn’t just be a finite number of them, or we could just drop them from the sequence. No, there would have to be an infinite subsequence of non-integrable $f_n$ , which would mean an infinite subsequence of their integrals would diverge to $\infty$ . Thus when we take the limit inferior of the integrals we get $\infty$ , as we do for the integral of $f$ itself, and the inequality still holds.

June 16, 2010 Posted by John Armstrong | Analysis, Measure Theory | 6 Comments

The Monotone Convergence Theorem

We want to prove a strengthening of the dominated convergence theorem. If $\{f_n\}$ is an a.e. increasing sequence of extended real-valued, non-negative, measurable functions, and if $f_n$ converges to $f$ pointwise a.e., then

$\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\int f\,d\mu$

If $f$ is integrable, then $f$ dominates the sequence $\{f_n\}$ , and so the dominated convergence theorem itself gives us the result we assert. What we have to show is that if $\int f\,d\mu=\infty$ , then the limit diverges to infinity. Or, contrapositively, if the limit doesn’t diverge then $f$ must be integrable.

But this is the limit of a sequence of real numbers, and so if it converges then it’s Cauchy. That is, we can conclude that

$\displaystyle\lim\limits_{m,n\to\infty}\left\lvert\int f_m\,d\mu-\int f_n\,d\mu\right\rvert=0$

Our assumption that $f_n$ is a.e. increasing tells us that for any fixed $m$ and $n$ , the difference $f_m-f_n$ is either a.e. non-negative or a.e. non-positive. That is,

$\displaystyle\left\lvert\int f_m\,d\mu-\int f_n\,d\mu\right\rvert=\int\lvert f_m-f_n\rvert\,d\mu$

And thus the sequence $\{f_n\}$ is mean Cauchy, and thus mean convergent to some integrable function $g$ , which must be equal to $f$ almost everywhere.

One nice use of this is when talking about series of functions. If $\{f_n\}$ is a sequence of integrable functions so that

$\displaystyle\sum\limits_{n=1}^\infty\int\lvert f_n\rvert\,d\mu<\infty$

then I say that the series

$\displaystyle\sum\limits_{n=1}^\infty f_n(x)$

converges a.e. to an integrable function $f$ , and further that

$\displaystyle\int f\,d\mu=\int\sum\limits_{n=1}^\infty f_n\,d\mu=\sum\limits_{n=1}^\infty\int f_n\,d\mu$

To see this, we let $S_n(x)$ be the partial sum

$\displaystyle S_n(x)=\sum\limits_{i=1}^n\lvert f_i(x)\rvert$

which gives us a pointwise increasing sequence of non-negative measurable functions. The monotone convergence theorem tells us that these partial sums converge pointwise to some $S(x)$ and that

$\displaystyle\int S\,d\mu=\lim\limits_{n\to\infty}\int S_n\,d\mu=\lim\limits_{n\to\infty}\int\sum\limits_{i=1}^n\lvert f_i\rvert\,d\mu=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\int\lvert f_i\rvert\,d\mu=\sum\limits_{i=1}^\infty\int\lvert f_i\rvert\,d\mu$

But this is exactly the sum we assumed to converge before. Thus the function $S$ is integrable and the series of the $f_n$ is absolutely convergent. That is, since $S$ must be a.e. finite, the series

$\displaystyle\sum\limits_{n=1}^\infty f_n(x)$

is absolutely convergent for almost all $x$ , and so it must be convergent pointwise almost everywhere. Since $S$ dominates the partial sums

$\displaystyle S(x)=\sum\limits_{i=1}^\infty\lvert f_i(x)\rvert\geq\sum\limits_{i=1}^n\lvert f_i(x)\rvert\geq\left\lvert\sum\limits_{i=1}^nf_i(x)\right\rvert$

the bounded convergence theorem tells us that limits commute with integrations here, and thus that

$\displaystyle\begin{aligned}\int f\,d\mu&=\int\sum\limits_{i=1}^\infty f_i\,d\mu\\&=\int\lim\limits_{n\to\infty}\sum\limits_{i=1}^nf_i\,d\mu\\&=\lim\limits_{n\to\infty}\int\sum\limits_{i=1}^nf_i\,d\mu\\&=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\int f_i\,d\mu\\&=\sum\limits_{i=1}^\infty\int f_i\,d\mu\end{aligned}$

June 15, 2010 Posted by John Armstrong | Analysis, Measure Theory | 10 Comments

The Integral Mean Value Theorem

We have an analogue of the integral mean value theorem that holds not just for single integrals, not just for multiple integrals, but for integrals over any measure space.

If $f$ is an essentially bounded measurable function with $\alpha\leq f\leq\beta$ a.e. for some real numbers $\alpha$ and $\beta$ , and if $g$ is any integrable function, then there is some real number $\gamma$ with $\alpha\leq\gamma\leq\beta$ so that

$\displaystyle\int f\lvert g\rvert\,d\mu=\gamma\int\lvert g\rvert\,d\mu$

Actually, this is a statement about finite measure spaces; the function $g$ is here so that the indefinite integral of $\lvert g\rvert$ will give us a finite measure on the measurable space $X$ to replace the (possibly non-finite) measure $\mu$ . This explains the $g$ in the multivariable case, which wasn’t necessary when we were just integrating over a finite interval in the one-variable case.

Okay, we know that $\alpha\leq f\leq\beta$ a.e., and so $\alpha\lvert g\rvert\leq f\lvert g\rvert\leq\beta\lvert g\rvert$ a.e. as well. This tells us that $f\lvert g\rvert$ is integrable. And thus we conclude

$\displaystyle\alpha\int\lvert g\vert\,d\mu\leq\int f\lvert g\rvert\,d\mu\leq\beta\int\lvert g\rvert\,d\mu$

Now either the integral of $\lvert g\rvert$ is zero or it’s not. If it’s zero, then $g$ is zero a.e., and so is $f\lvert g\rvert$ , and our assertion follows for any $\gamma$ we like. On the other hand, if it’s not we can divide through to find

$\displaystyle\alpha\leq\frac{\int f\lvert g\rvert\,d\mu}{\int\lvert g\vert\,d\mu}\leq\beta$

this term in the middle is our $\gamma$ .

June 14, 2010 Posted by John Armstrong | Analysis, Measure Theory | 5 Comments

Using the Dominated Convergence Theorem

Now that we’ve established Lebesgue’s dominated convergence theorem, we can put it to good use.

If $f$ is a measurable function and $g$ is an integrable function so that $\lvert f\rvert\leq g$ a.e., then $f$ is integrable. Indeed, we can break any function into positive and negative parts $f^+$ and $f^-$ , which themselves must satisfy $f^\pm\leq g$ a.e., and which are both nonnegative. So if we can establish the proposition for nonnegative functions the general case will follow.

If $f$ is simple, then it has to be integrable or else $g$ couldn’t be. In general, there is an increasing sequence $\{f_n\}$ of nonnegative simple functions converging pointwise to $f$ . Each of the $f_n$ is itself less than $g$ , and thus is integrable; we find ourselves with a sequence of integrable functions, dominated by the integrable function $g$ , converging pointwise to $f$ . The dominated convergence theorem then tells us that $f$ is integrable.

Next, a measurable function $f$ is integrable if and only if its absolute value $\lvert f\rvert$ is integrable. In fact, we already know that $f$ being integrable implies that $\lvert f\rvert$ is, but we can now go the other way. But then $f$ is a measurable function and $\lvert f\rvert$ an integrable function with $\lvert f\rvert\leq\lvert r\rvert$ everywhere. The previous result implies that $f$ is integrable.

If $f$ is integrable and $g$ is an essentially bounded measurable function, then the product $fg$ is integrable. Indeed, if $\lvert g\rvert\leq c$ a.e., then $\lvert fg\rvert\leq c\lvert f\rvert$ a.e. as well. Since $c\lvert f\rvert$ is integrable, our first result tells us that $\lvert fg\rvert$ is integrable, and our second result then tells us that $fg$ is integrable.

Finally, for today, if $f$ is an essentially bounded measurable function and $E$ is a measurable set of finite measure, then $f$ is integrable over $E$ . Since $E$ has finite measure, its characteristic function $\chi_E$ is integrable. Then our previous result tells us that $f\chi_E$ is integrable, which is what it means for $f$ to be integrable over $E$ .

June 11, 2010 Posted by John Armstrong | Analysis, Measure Theory | 1 Comment

Lebesgue’s Dominated Convergence Theorem

The dominated convergence theorem provides a nice tool to make sure certain sequences of integrable functions converge (in the mean) to integrable limits. Yes, we have the definition and the characterization in terms of convergence in measure, but this theorem is often easier to apply.

If $\{f_n\}$ is a sequence of integrable functions converging in measure or converging a.e. to a function $f$ , and if $g$ is an integrable function which “dominates” the sequence $\{f_n\}$ — we have $\lvert f_n(x)\rvert\leq g(x)$ for almost all $x$ — then $f$ is integrable and the sequence $\{f_n\}$ converges to $f$ in the mean. It’s important to note that we do not assume that $f$ is integrable; that’s part of the conclusion.

If we start by assuming that $\{f_n\}$ converges in measure to $f$ , then yesterday’s result immediately tells us that $\{f_n\}$ converges in the mean to $f$ ; the uniformity assumptions from that theorem are consequences of the inequalities

$\displaystyle\int\limits_E\lvert f\rvert\,d\mu\leq\int g\,d\mu$

Now, if we only assume that $\{f_n\}$ converges a.e. to $f$ , then we can reduce to convergence in measure by using $g$ . By throwing out a set of measure zero, we will assume that $\lvert f_n(x)\rvert$ and $\lvert f(x)\rvert$ are all no more than $g(x)$ everywhere. Then for every fixed positive $\epsilon$ we can write

$\displaystyle E_n=\bigcup\limits_{i=n}^\infty\left\{x\in X\big\vert\lvert f_i(x)-f(x)\rvert\geq\epsilon\right\}\subseteq\left\{x\in X\bigg\vert\lvert g(x)\rvert\geq\frac{\epsilon}{2}\right\}$

but the measure of this latter set is finite, and so $\mu(E_n)<\infty$ for all $n$ . A.e. convergence tells us that the measure of the intersection of all the $E_n$ is $0$ . By continuity, we conclude that

$\displaystyle\limsup\limits_{n\to\infty}\mu\left(\left\{x\in X\big\vert\lvert f_n(x)-f(x)\rvert\geq\epsilon\right\}\right)\leq\lim\limits_{n\to\infty}\mu(E_n)=\mu\left(\lim\limits_{n\to\infty}E_n\right)=0$

That is, in the presence of a dominating function $g$ , convergence a.e. implies convergence in measure, and thus implies convergence in mean.

Incidentally, since we know that

$\displaystyle\left\lvert\int f_n\,d\mu-\int f\,d\mu\right\rvert=\left\lvert\int f_n-f\,d\mu\right\rvert\leq\int\lvert f_n-f\rvert\,d\mu$

we can use convergence in the mean to control the right hand side, and thus get control over the left hand side. That is, we find that

$\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\int\lim\limits_{n\to\infty}f_n\,d\mu$

The dominated convergence theorem shows that the integral and the limit commute so long as the sequence is dominated by some integrable function.

It should be noted that the dominating function is essential. Indeed, let $X$ be the closed unit interval with Lebesgue measure, and let $E_n=\left(0,\frac{1}{n}\right)$ . Now we consider the sequence $\{n\chi_{E_n}\}$ which converges in measure to $0$ . However, there is no integrable dominating function; we find that

$\displaystyle\int\lvert n\chi_{E_n}\rvert\,d\mu=1$

and so the sequence cannot converge in the mean to $0$ .

June 10, 2010 Posted by John Armstrong | Analysis, Measure Theory | 9 Comments

Equicontinuity, Convergence in Measure, and Convergence in Mean

First off we want to introduce another notion of continuity for set functions. We recall that a set function $\nu$ on a class $\mathcal{E}$ is continuous from above at $\emptyset$ if for every decreasing sequence of sets $E_n\in\mathcal{E}$ with $\lim_nE_n=\emptyset$ we have $\lim_n\nu(E_n)=0$ . If $\{\nu_m\}$ is a sequence of set functions, then, we say the sequence is “equicontinuous from above at $\emptyset$ ” if for every sequence $\{E_n\}\subseteq\mathcal{E}$ decreasing to $\emptyset$ and for every $\epsilon>0$ there is some number $N$ so that if $n\geq N$ we have $\lvert\nu_m(E_n)\rvert<\epsilon$ . It seems to me, at least, that this could also be called “uniformly continuous from above at $\emptyset$ “, but I suppose equicontinuous is standard.

Anyway, now we can characterize exactly how convergence in mean and measure differ from each other: a sequence $\{f_n\}$ of integrable functions converges in the mean to an integrable function $f$ if and only if $\{f_n\}$ converges in measure to $f$ and the indefinite integrals $\nu_n$ of $\lvert f_n\rvert$ are uniformly absolutely continuous and equicontinuous from above at $\emptyset$ .

We’ve already shown that convergence in mean implies convergence in measure, and we’ve shown that convergence in mean implies uniform absolute continuity of the indefinite integrals. All we need to show in the first direction is that if $\{f_n\}$ converges in mean to $f$ , then the indefinite integrals $\{\nu_n\}$ are equicontinuous from above at $\emptyset$ .

For every $\epsilon>0$ we can find an $N$ so that for $n\geq N$ we have $\lVert f_n-f\rVert_1<\frac{\epsilon}{2}$ . The indefinite integral of a nonnegative a.e. function is real-valued, countably additive, and nonnegative, and thus is a measure. Thus, like any measure, it’s continuous from above at $\emptyset$ . And so for every sequence $\{E_m\}$ of measurable sets decreasing to $\emptyset$ there is some $M$ so that for $m\geq M$ we find

$\displaystyle\begin{aligned}\int\limits_{E_m}\lvert f_n-f\rvert\,d\mu&<\frac{\epsilon}{2}\\\int\limits_{E_m}\lvert f\rvert\,d\mu&<\frac{\epsilon}{2}\end{aligned}$

the first for all $n$ from $1$ to $N$ . Then if $m\geq M$ we have

$\displaystyle\lvert\nu_n(E_m)\rvert=\int\limits_{E_m}\lvert f_n\rvert\,d\mu\leq\int\limits_{E_m}\lvert f_n-f\rvert\,d\mu+\int\limits_{E_m}\lvert f\rvert\,d\mu<\epsilon$

for every positive $n$ . We control the first term in the middle by the mean convergence of $\{f_n\}$ for $n\geq N$ and by the continuity from above of $\int_E\lvert f_n-f\rvert\,d\mu$ for $n\leq N$ . And so the $\nu_n$ are equicontinuous from above at $\emptyset$ .

Now we turn to the sufficiency of the conditions: assume that $\{f_n\}$ converges in measure to $f$ , and that the sequence $\{\nu_n\}$ of indefinite integrals is both uniformly absolutely continuous and equicontinuous from above at $\emptyset$ . We will show that $\{f_n\}$ converges in mean to $f$ .

We’ve shown that $N(f_n)=\{x\in X\vert f(x)\neq0\}$ is $\sigma$ -finite, and so the countable union

$E_0=\bigcup\limits_{n=1}^\infty N(f_n)$

of all the points where any of the $f_n$ are nonzero is again $\sigma$ -finite. If $\{E_n\}$ is an increasing sequence of measurable sets with $\lim_nE_n=E_0$ , then the differences $F_n=E_0\setminus E_n$ form a decreasing sequence $\{F_n\}$ converging to $\emptyset$ . Equicontinuity then implies that for every $\delta>0$ there is some $k$ so that $\nu_n(F_k)<\frac{\delta}{2}$ , and thus

$\displaystyle\int\limits_{F_k}\lvert f_m-f_n\rvert\,d\mu\leq\int\limits_{F_k}\lvert f_m\rvert\,d\mu+\int\limits_{F_k}\lvert f_n\rvert\,d\mu=\nu_m(F_k)+\nu_n(F_k)\leq\frac{\delta}{2}+\frac{\delta}{2}=\delta$

For any fixed $\epsilon>0$ we define

$\displaystyle G_{mn}=\left\{x\in X\big\vert\lvert f_m-f_n\rvert\geq\epsilon\right\}$

and it follows that

$\displaystyle\begin{aligned}\int\limits_{E_k}\lvert f_m-f_n\rvert\,d\mu&\leq\int\limits_{E_k\cap G_{mn}}\lvert f_m-f_n\rvert\,d\mu+\int\limits_{E_k\setminus G_{mn}}\lvert f_m-f_n\rvert\,d\mu\\&\leq\int\limits_{E_k\cap G_{mn}}\lvert f_m-f_n\rvert\,d\mu+\epsilon\mu(E_k)\end{aligned}$

By convergence in measure and uniform absolute continuity we can make the integral over $E_k\cap G_{mn}$ arbitrarily small by choosing $m$ and $n$ sufficiently large. We deduce that

$\displaystyle\limsup\limits_{m,n\to\infty}\int\limits_{E_k}\lvert f_m-f_n\rvert\,d\mu\leq\epsilon\mu(E_k)$

and, since $\epsilon>0$ was arbitrary, we conclude

$\displaystyle\lim\limits_{m,n\to\infty}\int\limits_{E_k}\lvert f_m-f_n\rvert\,d\mu=0$

Now we can see that

$\displaystyle\int\lvert f_m-f_n\rvert\,d\mu=\int\limits_{E_0}\lvert f_m-f_n\rvert=\int\limits_{E_k}\lvert f_m-f_n\rvert+\int\limits_{F_k}\lvert f_m-f_n\rvert\,d\mu$

and thus

$\displaystyle\limsup\limits_{m,n\to\infty}\int\lvert f_m-f_n\rvert\,d\mu<\delta$

and, since $\delta>0$ is arbitrary,

$\displaystyle\lim\limits_{m,n\to\infty}\int\lvert f_m-f_n\rvert\,d\mu=0$

That is, the sequence $\{f_n\}$ is Cauchy in the mean. But we know that the $L^1$ norm is complete, and so $\{f_n\}$ converges in the mean to some function $g$ . But this convergence in mean implied convergence in measure, and so $\{f_n\}$ converges in measure to $g$ , and thus $f=g$ almost everywhere.

June 9, 2010 Posted by John Armstrong | Analysis, Measure Theory | 3 Comments

Mean Convergence is Complete

I evidently have to avoid saying $L^1$ in post titles because WordPress’ pingbacks can’t handle the unicode character ¹…

Anyhow, today we can show that the $L^1$ norm gives us a complete metric structure on the space of all integrable functions on $X$ . But first:

If $\{f_n\}$ is a mean Cauchy sequence of integrable simple functions converging in measure to $f$ , then it converges in the mean to $f$ as well. Indeed, for any fixed $m$ the sequence $\{\lvert f_n-f_m\rvert\}$ is a mean Cauchy sequence of integrable simple functions converging in measure to $\lvert f-f_m\rvert$ . Thus we find

$\displaystyle\int\lvert f-f_m\rvert\,d\mu=\lim\limits_{n\to\infty}\int\vert f_n-f_m\rvert\,d\mu$

But the fact that $\{f_n\}$ is mean Cauchy means that the integral on the right gets arbitrarily small as we take large enough $m$ and $n$ . And so on the left, the integral gets arbitrarily small as we take a large enough $m$ . That is,

$\displaystyle\lim\limits_{m\to\infty}\int\vert f-f_m\rvert\,d\mu=0$

and $\{f_n\}$ converges in the mean to $0$ .

As a quick corollary, given any integrable function $f$ and positive number $\epsilon$ , there is some integrable simple function $g$ with $\lVert f-g\rVert_1<\epsilon$ . Indeed, since $f$ is integrable there must be some sequence $\{f_n\}$ of integrable simple functions converging in measure to it, and we can pick $g=f_n$ for some sufficiently large $n$ .

Now, if $\{f_n\}$ is a mean Cauchy sequence of integrable functions — any integrable functions — then there is some integrable function $f$ to which the sequence converges in the mean. The previous corollary tells us that for every $f_n$ there’s some integrable simple $g_n$ so that $\lVert f_n-g_n\rVert_1<\frac{1}{n}$ . This gives us a new sequence $\{g_n\}$ , which is itself mean Cauchy. Indeed, for any $\epsilon>0$ choose an $N>\frac{3}{\epsilon}$ large enough so that $\lVert f_n-f_m\rVert_1<\frac{\epsilon}{3}$ . Then for $n,m\geq N$ we have

$\displaystyle\lVert g_n-g_m\rVert_1\leq\lVert g_n-f_n\rVert_1+\lVert f_n-f_m\rVert_1+\lVert f_m-g_m\rVert_1\leq\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$

Since $\{g_n\}$ is Cauchy in measure it converges in measure to some function $f$ . Then our first result shows that $\{g_n\}$ also converges in the mean to $f$ — that $\lVert g_n-f\rVert_1$ goes to zero as $n$ goes to $\infty$ . But we can also write

$\displaystyle\lVert f_n-f\rVert_1\leq\lVert f_n-g_n\rVert_1+\lVert g_n-f\rVert_1\leq\frac{1}{n}+\lVert g_n-f\rVert_1$

so if we choose a large enough $n$ , this will get arbitrarily small as well. That is, $\{f_n\}$ also converges in the mean to $f$ .

This shows that any mean Cauchy sequence of integrable functions is also mean convergent to some function, and thus the space of integrable functions equipped with the $L^1$ norm is a Banach space.

June 8, 2010 Posted by John Armstrong | Analysis, Measure Theory | 5 Comments

When is an Integral Zero?

We’ve got some interesting results about when integrals come out to be zero.

First up: if $f$ is an a.e. nonnegative integrable function, then $\int f\,d\mu=0$ if and only if $f=0$ almost everywhere. In one direction, if $f=0$ a.e. then we can pick all $f_n=0$ as a sequence of simple functions converging in measure to $f$ ; clearly the limit of their integrals is zero. On the other hand if $\{f_n\}$ a mean Cauchy sequence of integrable simple functions converging in measure to $f$ , then so does $\{\lvert f_n\rvert\}$ since $f$ is a.e. nonnegative. If $\int f\,d\mu=0$ then we must have

$\displaystyle\lim\limits_{n\to\infty}\int\lvert f_n\rvert\,d\mu=0$

that is, $\{f_n\}$ converges in mean to $0$ , which we know means it converges in measure to $0$ as well. But we picked this sequence to converge in measure to $f$ , and so $f=0$ almost everywhere.

As a quick corollary, if $f$ is integrable and $E$ is a set of measure zero, then $\int_Ef\,d\mu=0$ , because this is really the integral of $f\chi_E$ , which is a.e. zero.

A sort of a converse: if $f$ is integrable and positive a.e. on a measurable set $E$ , and if $\int_Ef\,d\mu=0$ , then $\mu(E)=0$ . We define $F_0=\{x\in X\vert f(x)>0\}$ and

$\displaystyle F_n=\left\{x\in X\bigg\vert f(x)\geq\frac{1}{n}\right\}$

for all positive integers $n$ ; our assumption tells us that $E\setminus F_0$ has measure zero, so if we can show that $E\cap F_0$ does too, then we’ll see that $\mu(E)=0$ . But we have

$\displaystyle0\leq\frac{1}{n}\mu(E\cap F_n)\leq\int\limits_{E\cap F_n}f\,d\mu\leq\int\limits_Ef\,d\mu=0$

where the last inequality holds because $f\geq0$ a.e. on $E\setminus F_n$ . This shows that $\mu(E\cap F_n)=0$ for all $n$ . But $F_0$ is the union of all the $F_n$ , and so

$\displaystyle\mu(E\cap F_0)\leq\sum\limits_{i=1}^\infty\mu(E\cap F_n)=0$

as we wanted to show.

Now if $f$ is integrable and $\int_Ff\,d\mu=0$ for every measurable $F$ , then $f=0$ almost everywhere. Indeed, if $E=\{x\in X\vert f(x)>0\}$ , then our hypothesis says that $\int_Ef\,d\mu=0$ , and the previous result then shows that $\mu(E)=0$ . Applying the same reasoning to $-f$ shows that the same is true of the set of points where $f$ is negative.

Finally, we can show that if $f$ is integrable, then the set $N(f)=\{x\in X\vert f(x)\neq0\}$ is $\sigma$ -finite. To see this, let $\{f_n\}$ be a mean Cauchy sequence of integrable simple functions converging in measure to $f$ . For every $n$ , $N(f_n)$ is a measurable set of finite measure. Now set

$\displaystyle E=N(f)\setminus\bigcup\limits_{n=1}^\infty N(f_n)$

If $F$ is any measurable subset of $E$ , then we can see that

$\displaystyle\int\limits_Ff\,d\mu=\lim\limits_{n\to\infty}\int\limits_Ff_n\,d\mu=0$

since $F$ is outside each $N(f_n)$ . Now our previous result shows us that $f=0$ a.e. on $E$ , but since $E\subseteq N(f)$ we have $f\neq0$ everywhere in $E$ ! The only possibility is that $E$ itself has measure zero. And thus

$\displaystyle N(f)\subseteq E\cup\bigcup\limits_{n=1}^\infty N(f_n)$

writes $N(f)$ as the (countable) union of a bunch of sets of finite measure, as desired.

June 7, 2010 Posted by John Armstrong | Analysis, Measure Theory | 4 Comments

Mean Convergence Properties of Integrals

Okay, we’ve got our general definition of integrable functions, and we’ve reestablished a bunch of our basic properties in this setting. Let’s consider some properties that involve the $L^1$ norm.

First off, the basic definitions of the $L^1$ norm carries across: we set

$\displaystyle\lVert f\rVert_1=\int f\,d\mu$

and we say that a sequence $\{f_n\}$ of integrable functions is Cauchy in the mean or is mean Cauchy if $\lVert f_n-f_m\rVert_1$ goes to zero for sufficiently large $m$ and $n$ . We immediately find that any sequence that is Cauchy in the mean is also Cauchy in measure, since our earlier proof didn’t really depend on the functions being simple.

We also have a couple more results whose proofs don’t really depend on the simplicity of $f_n$ , and so these carry across without change. If $\{f_n\}$ is mean Cauchy, with indefinite integrals $\{\nu_n\}$ , then the collection of set functions $\{\nu_n\}$ is uniformly absolutely continuous. Further, we can define the limiting set function

$\displaystyle\nu(E)=\lim\limits_{n\to\infty}\nu_n(E)$

for every measurable $E\subseteq X$ , and we find that this set function $\nu$ is finite-valued and countably additive.

So now we can formally define the obvious notion: the sequence $\{f_n\}$ of integrable functions “converges in the mean” or is “mean convergent” to an integrable function $f$ if $\lVert f_n-f\rVert_1$ goes to zero as $n$ goes to $\infty$ . And, as we might expect, if $\{f_n\}$ converges in the mean to $f$ then it converges in measure as well.

Indeed, for every $\epsilon>0$ we define

$\displaystyle E_n=\left\{x\in X\big\vert\lvert f_n-f\rvert\geq\epsilon\right\}$

and then we see that

$\displaystyle\int\lvert f_n-f\rvert\,d\mu\geq\int\limits_{E_n}\lvert f_n-f\rvert\,d\mu\geq\epsilon\mu(E_n)$

Thus $\mu(E_n)$ must go to $0$ as $n\to\infty$ , and so $\{f_n\}$ converges in measure to $f$ .

June 4, 2010 Posted by John Armstrong | Analysis, Measure Theory | 3 Comments

Properties of Integrable Functions

Now we’ve got a definition of the integral of a wider variety of functions than before, so let’s look over the basic properties.

First of all, from what we know about convergence in measure and algebraic and order properties of integrals of simple functions, we can see that if $f$ and $g$ are integrable functions and $\alpha$ is a real number, then so are the absolute value $\lvert f\rvert$ , the scalar multiple $\alpha f$ , and the sum $f+g$ . As special cases, we can see that the positive and negative parts

$\displaystyle\begin{aligned}f^+&=\frac{1}{2}(\lvert f\rvert+f)\\f^-&=\frac{1}{2}(\lvert f\rvert-f)\end{aligned}$

are both integrable.

If $E$ is a measurable set and $\{f_n\}$ is a mean Cauchy sequence of integrable simple functions converging in measure to $f$ , then it should be clear that $\{f\chi_E\}$ is mean Cauchy and converges in measure to $f\chi_E$ . Thus we can define

$\displaystyle\int\limits_Ef\,d\mu=\int f\chi_E\,d\mu$

Since integrals of simple functions are linear, and limits are linear, we immediately conclude that

$\displaystyle\int\alpha f+\beta g\,d\mu=\alpha\int f\,d\mu+\beta\int g\,d\mu$

as before, but now for general integrable functions. Similarly, if $f$ is nonnegative a.e., we can find a sequence of nonnegative simple functions converging a.e. (and thus in measure) to $f$ . The integral of each of these functions is nonnegative, and so their limit must be as well. That is, if $f\geq0$ a.e., then

$\displaystyle\int f\,d\mu\geq0$

Now, all our properties that we proved using only these two linearity and order properties follow. If $f\geq g$ a.e., then

$\displaystyle\int f\,d\mu\geq\int g\,d\mu$

For any two integrable functions $f$ amd $g$ we find

$\displaystyle\int\lvert f+g\rvert\,d\mu\leq\int\lvert f\rvert\,d\mu+\int\lvert g\rvert\,d\mu$

and

$\displaystyle\left\lvert\int f\,d\mu\right\rvert\leq\int\lvert f\rvert\,d\mu$

If $\alpha$ and $\beta$ are real numbers so that $\alpha\leq f(x)\leq\beta$ for almost all $x\in E$ , then we have

$\displaystyle\alpha\mu(E)\leq\int\limits_Ef\,d\mu\leq\beta\mu(E)$

and if an integrable function $f$ is a.e. nonnegative, then its indefinite integral is monotone.

It takes a bit of work, though, to check that the indefinite integral $\nu$ of an integrable function $f$ is absolutely continuous. If $\{f_n\}$ is a mean Cauchy sequence converging in measure to $f$ , then we have

$\displaystyle\left\lvert\int\limits_Ef\,d\mu\right\rvert\leq\left\lvert\int\limits_Ef_n\,d\mu\right\rvert+\left\lvert\int\limits_Ef\,d\mu-\int\limits_Ef_n\,d\mu\right\rvert$

We know that the indefinite integrals $\nu_n$ of the $f_n$ are uniformly absolutely continuous, so we have control over the size of the first term on the right. The second term on the right can be kept small by choosing a large enough $n$ , since $\{f_n\}$ converges in measure to $f$ .

Finally, the indefinite integral $\nu$ of $f$ is countably additive; since $\{f_n\}$ is a mean Cauchy sequence of simple functions we can write $\nu$ as the limit $\nu(E)=\lim_n\nu_n(E)$ , and we know that this limit is countably additive.

June 3, 2010 Posted by John Armstrong | Analysis, Measure Theory | 3 Comments

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