# Measurability of a function

I am having a hard time understanding a condition in building a measurable function. This is an example of a measurable function from this book (p. 55). Quoting verbatim:

Consider $R_n(\omega)$, the Rademacher functions. These are measurable because they are piecewise constant; namely, for any subset $A$ of $\mathbb{R} \cup \{+\infty\} \cup \{-\infty\}$, we have that $\{\omega \in I; R_n(\omega) \in A\}$ is a finite union of intervals.

where $I = [0,1]$ and the $n$th Rademacher function \begin{align} R_n(\omega) = \begin{cases} 1 & \textrm{if } a_n = 1 \newline -1 & \textrm{if } a_n = 0 \end{cases} \end{align} and $a_n$ is the $n$th number in the binary expansion of $\omega \in I$.

Namely, I don't get the condition placed regarding

any subset $A$ of $\mathbb{R} \cup \{+\infty\} \cup \{-\infty\}$

If the only values $R_n$ can take are +1 and -1, what's the point in saying that $A$ can be any interval on the extended real line? Isn't it just as valid to state

$\{\omega \in I; R_n(\omega) \in \{-1,1\}\}$

and argue that the Rademacher functions are measurable?

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You did not quote verbatim: the sentence in your first quote is missing and ending... – Mariano Suárez-Alvarez Feb 28 '11 at 1:27
@Mariano Oops! It's been fixed. – JasonMond Feb 28 '11 at 1:45

If $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces, then the definition of measurable function says that $f\colon X\to Y$ is measurable if and only if $f^{-1}(B) \in \mathcal{A}$ for every $B\in\mathcal{B}$.

(This is very similar/essentially identical to the definition of continuity for a function between topological spaces: if $(X,\tau)$ and $(Y,\sigma)$ are topological spaces, then $f\colon X\to Y$ is continuous if and only if $f^{-1}(B)\in\tau$ for every $B\in\sigma$.)

In general, if you have a $B\in\mathcal{B}$ such that $B\cap\mathrm{Im}(f)=\emptyset$, then you will automatically get that $f^{-1}(B)\in\mathcal{A}$; and if $\mathrm{Im}(f)\subseteq B$, then you will also automatically get $f^{-1}(B)\in \mathcal{A}$. So generally you are only concerned with elements of $\mathcal{B}$ that intersect, but don't contain, the image of $f$; not because you don't need to check the others, but rather because the other elements of $\mathcal{B}$ will trivially satisfy the condition.

What they are showing is that the Rademacher function is measurable when mapping into $\mathbb{R}$ with any $\sigma$-algebra, by showing that it is measurable relative to the $\sigma$-algebra $\mathcal{P}(\mathbb{R})$ of all subsets of $\mathbb{R}$. This is pretty much the strongest possible statement about measurability of these functions. In point of fact, it would be enough to consider only the sets $A=\{1\}$ and $A=\{-1\}$; any $A$ that contains both will necessarily have inverse image $I$ (measurable); any set that contains neither will necessarily have inverse image $\emptyset$ (measurable); and any set that contains exactly one of them will have inverse image equal to $R_n^{-1}(\{1\})$ or $R_n^{-1}(\{-1\})$ (depending on which one they contain). But there is no need to argue that way, and we can simply take an arbitrary subset $A$, even though the argument will just reduce to those two nontrivial cases.

What you cannot do is simply restrict to the case $A=\{1,-1\}$; this does not show that $R_n$ is measurable, because it does not show that a measurable subset of $\mathbb{R}$ which contains $1$ but not $-1$ has an inverse image in the (Borel) $\sigma$-algebra of $I$. That is, what you propose would be to check merely that $f^{-1}(B)$, where $B$ contains both $1$ and $-1$, is measurable. This is not sufficient to establish measurability.

(What you could do is talk about $A$ being any subset of $\{-1,1\}$, instead of saying that $A$ is any subset of $\mathbb{R}\cup\{+\infty\}\cup\{-\infty\}$, because the former suffice in this case.)

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Thanks! It makes a lot more sense now. – JasonMond Feb 28 '11 at 1:43