# The Unapologetic Mathematician

## 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 »

1. […] using our usual technique we […]

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2. […] 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 and are […]

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