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If $f : X \to \mathbb{R}^n$ is measurable, then $\langle f,f\rangle = ||f||_2^2: X \to \mathbb{R}$ is measurable (if $\langle f,f\rangle < c$ for $c > 0$ then $f$ should lie in an open ball with the origin as center and radius $\sqrt{c}$. The pre-image of which is measurable).

What can be said about $\langle f,z\rangle$ where $z \in \mathbb{R}^n$ is an arbitrary vector? Is it still measurable?

(A general query: I cannot access this website from my work place, where can I get help?)

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    What do you mean that you can't access this site from your work place? Can you access any of the stackexchange sites? Does it begin to load, but never finish? Is it possible that your work prevents you from going to some sites?2011-08-06
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    I cant access any stackexchange sites. We have a proxy server but mathoverflow works. It begins to load but never finishes as you said.2011-08-06
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    Well, $\langle \cdot, z\rangle$ is continuous, hence Borel measurable and $f$ has the property that pre-images of Borel sets are measurable by hypothesis.2011-08-06
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    Theo: Thanks, I think I understand now.2011-08-06
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    Do we automatically assume Borel measurability when we mention measurable? Otherwise, if I understood well, composition of continuous with Lebesgue-measurable is not necessarily measurable , if we have fog and f is continuous and g is just Lebesgue measurable.2011-08-06
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    @gary: It depends on which way around you compose: $h = g \circ f$ is measurable if $g$ is Borel and $f$ is Lebesgue measurable: you want to know if $h^{-1}(B)$ is measurable for $B$ Borel. But $h^{-1}(B) = f^{-1}(g^{-1}B)$ is the pre-image of a Borel set under $f$ hence is measurable. The other way around you can get into trouble: if $g$ is Lebesgue measurable then $L = g^{-1}(B)$ is only Lebesgue measurable and you don't know anything about $f^{-1}(L)$ of a Lebesgue measurable set $L$, so even if $f$ is continuous the resulting map $h = g \circ f$ need not be measurable.2011-08-06
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    gary: Thanks for the note (I could finally understand the difference between Borel and Lebesuge measurability). I have assumed Borel measurability. However, as far as my understanding goes, if one cannot put a measure on all Borel sets, why is Borel measurability so important?2011-08-06
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    @jpv: if X is an abstract space (specially not metric/metrizable), talking about balls, and radius may not apply. And even if X is metric, there may be no equivalent of the origin defined. Are you assuming X=$\mathbb R^n$?2011-08-06
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    @gary: No, I made a blunder. I wanted to mean that if $\langle f,f \rangle < c$ then $f$ has to lie in that open ball. As the pre-image of an open ball is measurable, we have $\langle f,f \rangle$ to be measurable. I will correct my question.2011-08-06

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There seems to be some confusion here, so let me answer instead of commenting:

If $f: (X,\Sigma_{X}) \to (Y,\Sigma_{Y})$ is a function between measurable spaces, it is called measurable if $f^{-1}(S) \in \Sigma_{X}$ for all $S \in \Sigma_{Y}$. So much for the general definition.

Now if $f: (X, \Sigma_{X}) \to Y$ is a map to some (say separable metric) space then measurability always means measurable with respect to the Borel $\sigma$-algebra $\mathcal{B}_{Y}$ on $Y$ unless otherwise specified.

The reason for this is simple: we want at least continuous maps to be measurable. Even for a continuous $f: [0,1] \to [0,1]$, the pre-image of a Lebesgue measurable set need not be Lebesgue measurable, and that should probably be reason enough for dismissing the notion of Lebesgue-Lebesgue measurable maps as hopelessly useless (at least from this perspective).

It is an interesting (but somewhat non-trivial) exercise to determine all maps $f: [0,1] \to [0,1]$ with the property that $f^{-1}(L)$ is Lebesgue measurable for all Lebesgue measurable sets $L$. I don't want to give too much away, but a small hint anyway: By a theorem of Vitali a subset $S$ of $[0,1]$ has the property that all its subsets are Lebesgue measurable if and only if $S$ has measure zero.


We can't compose Lebesgue measurable functions without thinking hard: if $f:[0,1] \to [0,1]$ is continuous and $g: [0,1] \to [0,1]$ is Lebesgue measurable in the sense that $L = g^{-1}(B)$ is Lebesgue measurable for all Borel sets $B$ then we're doomed: We simply can't say anything about $f^{-1}(L) = f^{-1}(g^{-1}(B))$ without knowing more about $f$ and $g$.

Fortunately, we're in the other situation: we're given that $f: (X,\Sigma_{X}) \to \mathbb{R}^{n}$ is measurable, so $f^{-1}(B) \in \Sigma_{X}$ for all Borel sets. We also have a Borel measurable (even continuous) map $g: \mathbb{R}^{n} \to \mathbb{R}$ and everything works out the way we want it to be: $g^{-1}(B)$ is Borel, hence $f^{-1}(g^{-1}(B)) \in \Sigma_X$, as we wanted.

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    Theo: Thanks for the nice answer.2011-08-06
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    I meant to say it is a nice answer to my not-so-great comment.2011-08-06