Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

I saw the following exercise:

Prove or give a counterexample: If $\{f_{i}\}_{i\in I}$ are continuous functions $f_{i}:\, X\to\mathbb{R}$ then ${\displaystyle \sup_{i\in\mathbb{I}}f_{i}}$ is measurable.

I think that this claim is false, if $\{f_{i}\}_{i\in I}$ are continuous functions $f_{i}:\, X\to\mathbb{R}$ then ${\displaystyle \sup_{i\in\mathbb{I}}f_{i}}$ is lower semi-continuous but necessarily upper semi-continuous, plus the index set can be uncountable which is a good source for a counterexample.

I tried taking $X=\mathbb{R}$ with the Borel $\sigma$-algebra and some non-measurable set $K$ and tried to define $f_{i}$ s.t $$f^{-1}((0,\infty))=\cup_{i\in\mathbb{R}}f_{i}^{-1}((0,\infty))=\cup_{i\in k}\{i\}=K$$ but I failed doing so (I can do this if $f_{i}$ are not continuous).

Can someone please help with this exercise?

share|cite|improve this question
Similar question showing that lower semicontinuous functions are measurable: Subset of the preimage of a semicontinuous real function is Borel – user52660 Dec 11 '12 at 17:11
The correct spelling is "continuous"; also "counterexample" is one word. I edited accordingly. – Nate Eldredge Dec 11 '12 at 17:45
@NateEldredge - thanks for the correction! – Belgi Dec 11 '12 at 18:20
Didn't you just post this question? – David Mitra Dec 11 '12 at 19:14
@DavidMitra - Yes, that post is the first part of a question I am trying to do (and I didn't want to ask about the other parts that I didn't try) and this is the third and last part of the question (I managed the second part doing something similir to what the answer to that post hinted for) – Belgi Dec 11 '12 at 19:20
up vote 1 down vote accepted

It depends on your $\sigma$-algebra on $X$. If you consider for example the trivial $\sigma$-algebra $\mathcal{A} := \{\emptyset,X\}$, then a continuous function $f:X \to \mathbb{R}$ is not necessarily measurable, in particular the supremum is not measurable. (For example $X=[0,1]$, $f(x) := x$.)

If you have some metric (or topology) on $X$ and you consider the Borel-$\sigma$-algebra (so the $\sigma$-algebra generated by the open sets) then it's true if $I$ is countable: Since $f_i: X \to \mathbb{R}$ are continuous, they are also measurable (pre-images of open sets are open!) and the supremum of (countable) measurable functions is again measurable. This follows from $$\{\sup_i f_i(x)>a\} = \bigcup_{i} \{f_i>a\}$$

share|cite|improve this answer

You must be assuming that continuous functions are measurable, hence I will assume that the $\sigma$-algebra under consideration contains the Borel sets. You already observed that $f$ is lower semi-continuous (but not necessarily upper semi-continuous). By definition, this means that $E_a = \{x \in X : f(x) \gt a\}$ is open in $X$, hence it is Borel measurable. This means precisely that $f$ is Borel measurable.

share|cite|improve this answer
What does "community wiki" mean above my name? – Daniel M. Dec 11 '12 at 22:46… – user27126 Dec 11 '12 at 23:17
Thanks. So now I know what it means but I don't know why my answer is community wiki... – Daniel M. Dec 11 '12 at 23:20

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.