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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?

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    Similar question showing that lower semicontinuous functions are measurable: [Subset of the preimage of a semicontinuous real function is Borel](http://math.stackexchange.com/q/79965)2012-12-11
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    The correct spelling is "continuous"; also "counterexample" is one word. I edited accordingly.2012-12-11
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    @NateEldredge - thanks for the correction!2012-12-11
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    Didn't you just post [this](http://math.stackexchange.com/questions/256301/proving-that-if-f-i-i-in-mathbbn-is-a-sequence-of-measurable-functio) question?2012-12-11
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    @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)2012-12-11
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    Well, unless I'm missing something, continuous functions are measurable; so, you could appeal to the first part.2012-12-11
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    @DavidMitra - but not the other way around, I gave an example of measurable functions that were not continuous...2012-12-11
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    Oh, sorry. You don't have a sequence of functions here. Silly me...2012-12-11
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    @DavidMitra - I think I have a counterexample, I will post it as CW.2012-12-11
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    The claim is false, take $X=\mathbb{R}$ and $I=\mathbb{R}$ and let $K\subseteq\mathbb{R}$ be unmeasurable set. for $i\not\in K$ define $f_{i}\equiv-1$ and for $i\in K$ define $f_{i}(x)$ to be a parabola with maximum at $x=i$ to be $f_{i}(i)=0$. Then every $f_{i}$ is continues but $f^{-1}([0,\infty))=\cup_{i\in\mathbb{R}}f_{i}^{-1}([0,\infty))=\cup_{i\in k}\{i\}=K$2012-12-11
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    had problems with posting this...2012-12-11
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    Hello, @dan. ${}{}$2012-12-12

2 Answers 2

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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 $$\{\sup_i f_i(x)>a\} = \bigcup_{i} \{f_i>a\}$$

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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.

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    What does "community wiki" mean above my name?2012-12-11
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    http://meta.stackexchange.com/questions/11740/what-are-community-wiki-posts2012-12-11
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    Thanks. So now I know what it means but I don't know why my answer is community wiki...2012-12-11