The Unapologetic Mathematician

Mathematics for the interested outsider

Integrable Functions

Now we can define what it means for a general real-valued function (not just a simple function) to be integrable: a function f is integrable if there is a mean Cauchy sequence of integrable simple functions \{f_n\} which converges in measure to f. We then define the integral of f to be the limit

\displaystyle\int f(x)\,d\mu(x)=\int f\,d\mu=\lim_{n\to\infty}\int f_n\,d\mu

But how do we know that this doesn’t depend on the sequence \{f_n\}?

We recall that we defined

\displaystyle N(f)=\{x\in X\vert f(x)\neq0\}

which must be measurable for any measurable function f. This is the only part of the space that matters when it comes to integrating f; clearly we can see that

\displaystyle\int f\,d\mu=\int\limits_{N(f)}f\,d\mu

since f is zero everywhere outside N(f).

Now, if both \{f_n\} and \{g_n\} converge in measure to f, then we can define E to be the (countable) union of all the N(f_n) and N(g_n). Just as clearly, we can see that

\displaystyle\begin{aligned}\int f_n\,d\mu&=\int\limits_Ef_n\,d\mu=\nu_n(E)\\\int g_n\,d\mu&=\int\limits_Eg_n\,d\mu=\lambda_n(E)\end{aligned}

where \nu_n is the indefinite integral of f_n, and \lambda_n is the indefinite integral of g_n. Then if we use \{f_n\} to define the integral of f we get

\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\lim\limits_{n\to\infty}\nu_n(E)=\nu(E)

while if we use \{g_n\} we get

\displaystyle\lim\limits_{n\to\infty}\int g_n\,d\mu=\lim\limits_{n\to\infty}\lambda_n(E)=\lambda(E)

But we know that since \{f_n\} and \{g_n\} both converge in measure to the same function, the limiting set functions \nu and \lambda coincide, and thus \nu(E)=\lambda(E). The value of the integral, then, doesn’t depend on the sequence of integrable simple functions!

June 2, 2010 Posted by | Analysis, Measure Theory | 5 Comments

Indefinite Integrals and Convergence II

Unlike our recent results, today’s proposition is specifically stated and proved for integrable simple functions, and won’t be generalized later.

If \{f_n\} and \{g_n\} are mean Cauchy sequences of integrable simple functions, then they’re both also Cauchy in measure, which implies that they each converge in measure to some function. If they converge to the same function (a.e.) f, then their indefinite integrals converge to the same limiting set function. That is, if \nu_n and \lambda_n are the indefinite integrals of f_n and g_n:

\displaystyle\begin{aligned}\nu_n(E)&=\int\limits_Ef_n\,d\mu\\\lambda_n(E)&=\int\limits_Eg_n\,d\mu\end{aligned}

then we can define the limiting functions

\displaystyle\begin{aligned}\nu(E)&=\lim\limits_{n\to\infty}\nu_n(E)\\\lambda(E)&=\lim\limits_{n\to\infty}\lambda_n(E)\end{aligned}

and we assert that \nu(E)=\lambda(E) for all measurable E\subseteq X.

For every \epsilon>0 and positive integer n we define the set

\displaystyle E_n=\left\{x\in X\big\vert\lvert f_n(x)-g_n(x)\rvert\geq\epsilon\right\}

And using our usual technique we find

\displaystyle E_n\subseteq\left\{x\in X\bigg\vert\lvert f_n(x)-f(x)\rvert\geq\frac{\epsilon}{2}\right\}\cup\left\{x\in X\bigg\vert\lvert f(x)-g_n(x)\rvert\geq\frac{\epsilon}{2}\right\}

Since \{f_n\} and \{g_n\} both converge in measure to f, the measures of both terms here go to zero as n gets large, and so \lim_n\mu(E_n)=0.

If E is a measurable set with \mu(E)<\infty, we have the inequality

\displaystyle\int\limits_E\lvert f_n-g_n\rvert\,d\mu\leq\int\limits_{E\setminus E_n}\lvert f_n-g_n\rvert\,d\mu+\int\limits_{E\cap E_n}\lvert f_n\rvert\,d\mu+\int\limits_{E\cap E_n}\lvert g_n\rvert\,d\mu

Here, the first term on the right is bounded above by \epsilon\mu(E). The other two terms can be made arbitrarily small by choosing a large enough n, by the uniform absolute continuity we showed yesterday. We can also see that

\displaystyle\lvert\nu_n(E)-\lambda_n(E)\rvert=\left\lvert\int_Ef_n\,d\mu-\int_Eg_n\,d\mu\right\rvert=\left\lvert\int_Ef_n-g_n\,d\mu\right\rvert\leq\int\limits_E\lvert f_n-g_n\rvert\,d\mu

and so it follows that \lim_n\lvert\nu_n(E)-\lambda_n(E)\rvert=0, and thus that \nu(E)=\lambda(E) for every measurable set E with finite measure. Since \nu and \lambda are countably additive, we immediately extend this result to all \sigma-finite sets E.

Okay, now here’s where our assumption really comes in: since each of the f_n and g_n is an integrable simple function, each one takes a nonzero value on a finite number of sets, each of which has a finite measure. Thus we can take the union E_0 of all these (countably many) sets, which is a \sigma-finite set. For any measurable set E, then, we have

\displaystyle\nu_n(E\setminus E_0)=\int\limits_{E\setminus E_0}f_n\,d\mu=0=\int\limits_{E\setminus E_0}g_n\,d\mu=\lambda_n(E\setminus E_0)

because all the functions f_n and g_n are identically zero off of E_0.

Therefore we conclude that \nu_n(E\setminus E_0)=0=\lambda_n(E\setminus E_0). This, then, implies that \nu(E)=\nu(E\cap E_0) and \lambda(E)=\lambda(E\cap E_0), and each of these sets is then \sigma-finite (as subsets of E_0). And so the proof is complete.

June 1, 2010 Posted by | Analysis, Measure Theory | 1 Comment

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