Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a locally-bounded function $f: W \rightarrow X$, $X \subseteq \mathbb{R}^n$, $W \subseteq \mathbb{R}^m$.

Assume that $f$ is Borel measurable

(for every open $O \in \Sigma_X$ ($\sigma$-algebra of $X$) we have $f^{-1}(O) \in \Sigma_W$ ($\sigma$-algebra of $W$))

Consider a locally-bounded discontinuous function $g: X \rightarrow Y$, $Y \subseteq \mathbb{R}^p$.

  1. Say if $\ g \circ f: W \rightarrow Y$ is measurable as well.

  2. If not: counterexample?

  3. If not: under which conditions is $g \circ f$ measurable?

share|cite|improve this question
It's obvious from the definition of measurability that if two functions are measurable then their composition is measurable. But at a wild guess I'd say $g$ is not necessarily measurable. – Ben Millwood Jun 29 '12 at 16:30
That's why I was first asking 1. :-) Seems to me that if we define $\bar{g}(w,x) := g(x)$, then for all $x \in X$, we have $w \mapsto \bar{g}(x,w)$ measurable (because constant). – Adam Jun 29 '12 at 16:39
...Why would you do that? What sort of thing is $w$? I don't even know what $w$ you are talking about in your question 1. – Ben Millwood Jun 29 '12 at 16:44
When I say "measurable", we mean functions from $W$ to $\mathbb{R}^{something}$. I'll adjust question 1. accordingly (to avoid confusion, I'll make it question 2.). So I don't see what do you mean by $g$ not measurable, because its domain is not ($\subseteq$) $W$. – Adam Jun 29 '12 at 16:51
I see what you mean now, but that's not the normal definition of measurable function. Usually, we define "measurable" to mean "the pre-images of measurable sets are measurable" (i.e. your definition, but using the appropriate Borel algebra for the domain and range of the function you are considering). Is it possible that you've misunderstood the definition you've been given? – Ben Millwood Jun 29 '12 at 16:57
up vote 1 down vote accepted

Consider $n=m=p=1$. Take $f:R \mapsto (0,1)$, $f(x) = \frac{1}{1+x^{2}}$, which is locally bounded, continuous and measurable. Let $V$ be the vitali set which is not measurable in $B(\mathcal{R})$. Let $g: (0,1) \mapsto \{0,1\}$,

$$g(x) = \left\{ \begin{array}{ll} 1 & \mbox{$x \in V$};\\ 0 & \mbox{$x \notin V$}.\end{array} \right. $$

$g \circ f$ is not measurable. If $g$ is continuous, $g \circ f$ is measurable. In general, if $g$ is measurable, $g \circ f$ is measurable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.