# The Unapologetic Mathematician

## 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.

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Posted by John Armstrong | Analysis, Measure Theory

## 6 Comments	»

1. [...] the function . If is a positive integer, we have the function . These are all continuous, which implies that they’re Borel measurable, and they send back to itself. We conclude that any positive [...]

Pingback by | May 4, 2010 | Reply

2. [...] continuous function such that if and only if . And since it’s continuous, it’s Borel measurable, and any Borel measurable function is Lebesgue [...]

Pingback by | May 5, 2010 | Reply

3. [...] now we turn to our convenient condition for measurability. Since we’ve handled the sets where and are infinite, we can assume that they’re [...]

Pingback by | May 7, 2010 | Reply

4. [...] now let’s use our convenient condition. Given a real number we know that if and only if for some . That is, we can [...]

Pingback by | May 10, 2010 | Reply

5. [...] is positive, is negative, and is a Hahn decomposition. Indeed, we know that and are measurable. Thus if is measurable, then is measurable, and we [...]

Pingback by | June 29, 2010 | Reply

6. [...] so that (or ) is a measurable set, then is a measurable function. We will show this using an equivalent definition of measurability — that will be measurable if we can show that for every real number the [...]

Pingback by | July 21, 2010 | Reply

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