More Measurable Real-Valued Functions

We want a few more convenient definitions of a measurable real-valued function. To begin with: a real-valued function $f$ on a measurable space $(X,\mathcal{S})$ is measurable if and only if for every real number $c$ the set $N(f)\cap\{x\in X\vert f(x)<c\}$ is measurable.

Indeed, say $f$ is measurable. If we take $M=\left(-\infty,c\right)$ then $M$ is a Borel set and $f^{-1}(M)=\{x\in X\vert f(x)<c\}$ . The measurability of $f$ tells us that $N(f)\cap f^{-1}(M)$ is measurable as a subset of $X$ .

Conversely, suppose that the given sets are all measurable. If $c_1\leq c_2$ are real numbers, then we can write $\left[c_1,c_2\right)=\left(-\infty,c_2\right)\setminus\left(-\infty,c_1\right)$ , and thus

$\displaystyle\{x\in X\vert c_1\leq f(x)<c_2\}=\{x\in X\vert f(x)<c_2\}\setminus\{x\in X\vert f(x)<c_1\}$

That is, if $M$ is any semiclosed interval then $N(f)\cap f^{-1}(M)$ is the difference of two measurable sets, and is thus measurable itself. If $\mathcal{M}$ is the collection of all the subsets $M\subseteq\mathbb{R}$ for which $N(f)\cap f^{-1}(M)$ is measurable, then $\mathcal{M}$ is a $\sigma$ -ring containing all semiclosed intervals. It must then contain all Borel sets, and so $f$ is measurable.

The same statement will hold true if we replace $<$ by $\leq$ , or by $>$ , or by $\geq$ . We walk through the exact same method as before, constructing left- or right-semiclosed intervals — and thus all Borel sets — from open or closed rays as needed.

In fact, we can even restrict $c$ to lie in some everywhere-dense subset $C\subseteq\mathbb{R}$ . For example, we might only check this condition for rational $c$ . Indeed, say we want to construct the closed interval $\left[a,b\right]$ . By density, we can find sequences $\{a_i\}$ (increasing) and $\{b_i\}$ (decreasing) of points in $C$ converging to $a$ and $b$ , respectively. Then we can construct the intervals $\left[a_i,b_i\right)$ or $\left(a_i,b_i\right]$ , and their intersection is the closed interval $\left[a,b\right]$ we want. Then the closed intervals generate the Borel sets, and we’re done.

All of these proofs, by the way, hinge on the fact that taking preimages and intersections commute with all of our set-theoretic constructions.

Now, if $f$ is a nonzero constant function $f(x)=c$ , then $f$ is measurable if and only if $X$ is a measurable subset of itself. Indeed, $N(f)=X$ , and $f^{-1}(M)$ is either $X$ or $\emptyset$ , according as $M$ does or does not contain $c$ . And since every $c$ must be contained in some measurable set, $X$ must be measurable for $f$ to be measurable.

More generally, the characteristic function $\xi_E$ of a set $E\subseteq X$ is measurable if and only if $E$ is a measurable subset of $X$ . This time, $N(f)=E$ , and $f^{-1}(M)$ is either $E$ or $\emptyset$ , according as $M$ contains $1$ or not.

If $f$ is a measurable function and $k$ is a nonzero real number, then the function $kf$ is also measurable. Indeed, it’s clear that $N(kf)=N(f)$ . We must check that $N(kf)\cap\{x\in X\vert kf(x)<c\}$ is measurable, but this set is equal to $N(f)\cap\{x\in X\vert f(x)<\frac{c}{k}\}$ , which is measurable.

Finally, every continuous function $f:\mathbb{R}\to\mathbb{R}$ is Borel measurable. Indeed, we can write any Borel set $B$ as a limit of open sets. The preimage of each open set is open, and thus Borel, and the preimage of the limit is the limit of the preimages, which is again Borel.

May 3, 2010 - Posted by John Armstrong | Analysis, Measure Theory

6 Comments »

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