The Unapologetic Mathematician

Mathematics for the interested outsider

Convergence in Measure and Algebra

Unlike our other methods of convergence, it’s not necessarily apparent that convergence in measure plays nicely with algebraic operations on the algebra of measurable functions. All our other forms are basically derived from pointwise convergence, and so the limit laws clearly hold; but it takes some work to see that the same is true for convergence in measure. So, for the rest of this post assume that \{f_n\} and \{g_n\} are sequences of finite-valued measurable functions converging in measure to f and g, respectively.

First up: if \alpha and \beta are real constants, then \{\alpha f_n+\beta g_n\} converges in measure to \alpha f+\beta g. Indeed, we find that

\displaystyle\begin{aligned}\lvert(\alpha f_n(x)+\beta g_n(x))-(\alpha f(x)+\beta g(x))\rvert&=\lvert\alpha(f_n(x)-f(x))+\beta(g_n(x)-g(x))\rvert\\&\leq\lvert\alpha\rvert\lvert f_n(x)-f(x)\rvert+\lvert\beta\rvert\lvert g_n(x)-g(x)\rvert\end{aligned}

Thus if \lvert f_n(x)-f(x)\rvert<\frac{\epsilon}{2\alpha} and \lvert g_n(x)-g(x)\rvert<\frac{\epsilon}{2\beta}, then \lvert(\alpha f_n(x)+\beta g_n(x))-(\alpha f(x)+\beta g(x))\rvert<\epsilon. That is,

\displaystyle\begin{aligned}&\left\{x\in X\big\vert\lvert(\alpha f_n(x)+\beta g_n(x))-(\alpha f(x)+\beta g(x))\rvert\geq\epsilon\right\}\\&\subseteq\left\{x\in X\bigg\vert\lvert f_n(x)-f(x)\rvert\geq\frac{\epsilon}{2\alpha}\right\}\cup\left\{x\in X\bigg\vert\lvert g_n(x)-g(x)\rvert\geq\frac{\epsilon}{2\beta}\right\}\end{aligned}

Since \{f_n\} and \{g_n\} converge in measure to f and g, we can control the size of each of these sets by choosing a sufficiently large n, and thus \{\alpha f_n+\beta g_n\} converges in measure to \alpha f+\beta g.

Next, if f=0 a.e., the sequence \{f_n^2\} converges in measure to f^2. Indeed, the set \left\{x\in X\big\vert\lvert f_n(x)-f(x)\rvert\geq\sqrt{\epsilon}\right\} differs negligibly from the set \left\{x\in X\big\vert\lvert f_n(x)\rvert\geq\sqrt{\epsilon}\right\}. This, in turn, is exactly the same as the set \left\{x\in X\big\vert\lvert f_n(x)^2\rvert\geq\epsilon\right\}, which differs negligibly from the set \left\{x\in X\big\vert\lvert f_n(x)^2-f(x)^2\rvert\geq\epsilon\right\}. Thus control on the measure of one of these sets is control on all of them.

Now we’ll add the assumption that the whole space X is measurable, and that \mu(X) is finite (that is, the measure space is “totally finite”). This will let us conclude that the sequence \{f_ng\} converges in measure to fg. As the constant c increases, the measurable set E_c=\left\{x\in X\big\vert\lvert g(x)\rvert\leq c\right\} gets larger and larger, while its complement X\setminus E_c gets smaller and smaller; this complement is measurable because X is measurable.

In fact, the measure of the complement must decrease to zero, or else we’d have some set of positive measure on which g(x) is larger than any finite c, and thus g(x)=\infty on a set of positive measure. But then \{g_n\} couldn’t converge to g in measure. Since \mu is totally finite, the measure \mu(X\setminus E_c) must start at some finite value and decrease to zero; if \mu(X) were infinite, these measures might all be infinite. And so for every \delta>0 there is some c so that \mu(X\setminus E_C)<\delta.

In particular, we can pick a c so that \mu(X\setminus E_c)<\frac{\delta}{2}. On E_c, then, we have \lvert f_ng-fg\rvert\leq\lvert f_n-f\rvert c. Convergence in measure tells us that we can pick a large enough n so that

\displaystyle\left\{x\in E_c\bigg\vert\lvert f_n-f\rvert\geq\frac{\epsilon}{c}\right\}

has measure less than \frac{\delta}{2} as well. The set \left\{x\in E_c\big\vert\lvert f_ng-fg\rvert\geq\epsilon\right\} must be contained between these two sets, and thus will have measure less than \delta for sufficiently large n.

Now we can show that \{f_n^2\} converges in measure to f^2 for any f, not just ones that are a.e. zero. We can expand (f_n-f)^2=f_n^2-2f_nf+f^2, and thus rewrite f_n^2=(f_n-f)^2+2f_nf-f^2. Our first result shows that \{f_n-f\} converges to f-f=0, and our second result then shows that \{(f_n-f)^2\} also converges to 0^2=0. Our third result shows that f_nf converges to f^2. We use our first result to put everything together again and conclude that \{(f_n-f)^2+2f_nf-f^2\} converges to f^2 as we asserted.

And finally we can show that \{f_ng_n\} converges in measure to fg. We can use the same polarization trick as we’ve used before. Write f_ng_n=\frac{1}{4}\left((f_n+g_n^2)-(f_n-g_n)^2\right); we’ve just verified that the squares converge to squares, and we know that linear combinations also converge to linear combinations, and so f_ng_n converges in measure to fg.

May 21, 2010 Posted by | Analysis, Measure Theory | 2 Comments

Convergence in Measure II

Sorry, I forgot to post this before I left this morning.

The proposition we started with yesterday shows us that on a set E of finite measure, a.e. convergence is equivalent to convergence in measure, and a sequence is Cauchy a.e. if and only if it’s Cauchy in measure. We can strengthen it slightly by removing the finiteness assumption, but changing from a.e. convergence to almost uniform convergence: almost uniform convergence implies convergence in measure. Indeed, if \{f_n\} converges to f almost uniformly then for any two positive real numbers \epsilon and \delta there is a measurable set F with \mu(F)<\delta and \lvert f_n(x)-f(x)\rvert<\epsilon for all x\in E\setminus F and sufficiently large n. Thus we can make the set F where f_n and f are separated as small as we like, as required by convergence in measure.

We also can show some common-sense facts about sequences converging and Cauchy in measure. First, if \{f_n\} converges in measure to f, then \{f_n\} is Cauchy in measure. We find that

\displaystyle\left\{x\in X\big\vert\lvert f_n(x)-f_m(x)\rvert\geq\epsilon\right\}\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_m(x)-f_m(x)\rvert\geq\frac{\epsilon}{2}\right\}

because if both f_n(x) and f_m(x) are within \frac{\epsilon}{2} of the same number f(x), then they’re surely within \epsilon of each other. And so if we have control on the measures of the sets on the right, we have control of the measure of the set on the left.

Secondly, if \{f_n\} also converges in measure to g, then it only makes sense that f and g should be “the same”. It wouldn’t do for a convergence method to have many limits for a convergent sequence. Of course, this being measure theory, “the same” means a.e. we have f(x)=g(x). But this uses almost the same relation:

\displaystyle\left\{x\in X\big\vert\lvert f(x)-g(x)\rvert\geq\epsilon\right\}\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_n(x)-g(x)\rvert\geq\frac{\epsilon}{2}\right\}

Since we can make each of the sets on the right arbitrarily small by choosing a large enough n, we must have \left\{x\in X\big\vert\lvert f(x)-g(x)\rvert\geq\epsilon\right\}=0 for every \epsilon>0; this implies that f=g almost everywhere.

Slightly deeper, if \{f_n\} is a sequence of measurable functions that is Cauchy in measure, then there is some subsequence \{f_{n_k}\} which is almost uniformly Cauchy. For every positive integer k we find some integer \bar{n}(k) so that if n,m\geq\bar{n}(k)

\displaystyle\mu\left(\left\{x\in X\bigg\vert\lvert f_n(x)-f_m(x)\rvert\geq\frac{1}{2^k}\right\}\right)<\frac{1}{2^k}

We define n_1=\bar{n}(1), and n_k to be the larger of n_{k-1}+1 or \bar{n}(k), to make sure that n_k is a strictly increasing sequence of natural numbers. We also define

\displaystyle E_k=\left\{x\in X\bigg\vert\lvert f_{n_k}(x)-f_{n_{k+1}}(x)\rvert\geq\frac{1}{2^k}\right\}

If k\leq i\leq j then for every x not in E_k\cup E_{k+1}\cup E_{k+2}\cup\dots we have

\displaystyle\lvert f_{n_i}(x)-f_{n_j}(x)\rvert\leq\sum\limits_{m=i}^\infty\lvert f_{n_m}(x)-f_{n_{m+1}}(x)\rvert<\frac{1}{2^{i-1}}

That is, the subsequence \{f_{n_k}\} is uniformly Cauchy on the set X\setminus\left(\cup_{i\geq k}E_i\right). But we also know that

\displaystyle\mu\left(\bigcup\limits_{m=k}^\infty E_m\right)\leq\sum\limits_{m=k}^\infty\mu(E_m)<\frac{1}{2^{k-1}}

and so \{f_{n_k}\} is almost uniformly Cauchy, as asserted.

Finally, we can take this subsequence that is almost uniformly Cauchy, and see that it must be a.e. Cauchy. We write f(x)=\lim_kf_{n_k}(x) at all x where this sequence converges. And then for every \epsilon>0,

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

The measure of the first set on the right is small for sufficiently large n and n_k by the assumption that \{f_n\} is Cauchy in measure. The measure of the second approaches zero because almost uniform convergence implies convergence in measure.

And thus we conclude that if \{f_n\} is Cauchy in measure, then there is some measurable function f(x) to which \{f_n\} converges in measure. The topology of convergence in measure may not come from a norm, but it is still complete.

May 21, 2010 Posted by | Analysis, Measure Theory | 4 Comments

   

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