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Let $f_1,..., f_n: \mathbb{R}\rightarrow\mathbb{R}$ be measurable functions. And $F:\mathbb{R}^n\rightarrow\mathbb{R}$ be continuous.

For $g:\mathbb{R}\rightarrow\mathbb{R}$, $g(x):=F(f_1(x),f_2(x),...,f_n(x))$. Is $g$ measurable?

I think I was be able to do that if F is measurable, but I don't know how is it if F is only continuous.

Awoo
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  • probably....... – mathworker21 Sep 19 '18 at 22:18
  • For you, is measurable Lebsegue-measurable or Borel-measurable? – qualcuno Sep 19 '18 at 22:21
  • I think I mean Lebsegue measurable – Awoo Sep 19 '18 at 22:22
  • Maybe it's better to ask: what is your definition of "measurable function"? – Alex Kruckman Sep 19 '18 at 22:24
  • $f^{-1}(I)$ is measurable for all interval $I$? – Awoo Sep 19 '18 at 22:26
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    @Awoo : Your interval definition corresponds to "Borel-measurable." Overall there can be ambiguity in the definition of "measurable function." For example, the definitions on these two links are different (and generally not the same unless we are working with the Borel sigma algebra): Link 1: http://mathworld.wolfram.com/MeasurableFunction.html and Link 2: https://en.wikipedia.org/wiki/Measurable_function – Michael Sep 20 '18 at 05:23
  • I agree with Michael. We should call a map from $\mathbb R$ to itself measurable if the inverse image of any Borel set is Borel. This is the universally accepted terminology in measure theory. If we want to consider other types of measurability we have to be specific. (After all Lebesgue measure is not the only measure!) – Kavi Rama Murthy Sep 20 '18 at 05:39

3 Answers3

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It is well known that there exist continuous functions $f : [0,2] \to \mathbb{R}$ that are not Lebesgue-measurable. Extend a function of this to the rest of the real line continuously, via

$$ h(x) = \cases{f(0) \quad x < 0 \\ f(x) \quad 0 \leq x \leq 2 \\ f(2) \quad x > 2} $$

Now, take $n = 1$, $f_1 \equiv id$, $F \equiv h$.

qualcuno
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  • But I think I am giving $f_i$ are measurable? Only $F$ is not necessary measurable. – Awoo Sep 19 '18 at 22:35
  • My bad, there's a simpler counterexample then. I'll edit. – qualcuno Sep 19 '18 at 22:36
  • You mean there could be "non-measurable part" in $F$ other than the giving $f_i$ part? – Awoo Sep 19 '18 at 22:39
  • In the latter example, $F$ is taken continuous but non measurable, and $f_1$ is trivially measurable. However, their composition is $F$, which is not measurable. – qualcuno Sep 19 '18 at 22:41
  • I think I was confusing myself. I don't think that function is not measurable. All real continuous function should be Lebesgue measurable, which gives $F$ should be also measurable, which lead to $h$ is also measurable – Awoo Sep 19 '18 at 23:42
  • I think Awoo is defining a measurable function $f$ to be a function such that for all $c \in \mathbb{R}$ we have ${x : f(x) > c}$ is measurable, as here: http://mathworld.wolfram.com/MeasurableFunction.html – Michael Sep 20 '18 at 04:13
  • @Michael yes, that's the definition I am using. Maybe he is using measurable to say 'borel measurable', in which case his claim is true because the composition is continuous and so preimages of borel will be borel. – qualcuno Sep 20 '18 at 04:44
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    The composition might not be continuous because the individual $f_i$ functions are not necessarily continuous. I think the definition "for all $c\in \mathbb{R}$ we have ${x: f(x)>c}$ is measurable" is generally different from "for all measurable subsets $Y$ of the range we have $f^{-1}(Y)$ is measurable in the domain," but is the same for Borel sigma-algebra. – Michael Sep 20 '18 at 04:57
  • You are correct, my bad. Preimages of open sets will be borel, but not necessarily open. So I'm not immediately sure if it holds when we take measureable to mean borel-measurable. What is certain is that this fails for Lebesgue-measurable. – qualcuno Sep 20 '18 at 05:02
  • My answer shows the property that ${x : g(x)>c}$ is measurable under any sigma-algebra. Under the Borel sigma-algebra, I believe this implies that $g^{-1}(Y)$ is measurable whenever $Y$ is measurable. It does not imply that in your Lebesgue example (which includes null sets). – Michael Sep 20 '18 at 05:10
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I think I was confusing myself. All real continuous function should be Lebesgue measurable, which gives $F$ should be also measurable, which lead to $h$ is also measurable.

Awoo
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Assume:

  • Let $\mathcal{X}$ be any set.

  • Let $\Sigma$ be any sigma-algebra on $\mathcal{X}$.

  • Fix $n$ as a positive integer.

  • For each $i \in \{1, ..., n\}$ and each $c \in \mathbb{R}$, the function $f_i:\mathcal{X}\rightarrow\mathbb{R}$ has the property that $\{x \in \mathcal{X} : f_i(x)>c\}$ is in $\Sigma$.

  • The function $F:\mathbb{R}^n\rightarrow \mathbb{R}$ is continuous.

Claim:

For each $c \in \mathbb{R}$, the set $\{x \in \mathcal{X} : F(f_1(x), ..., f_n(x)) > c\}$ is in $\Sigma$.

Proof:

Fix $c \in \mathbb{R}$. Define the set $\mathcal{Y} \subseteq\mathbb{R}^n$ by

$$ \mathcal{Y} = \{y \in \mathbb{R}^n : F(y) > c\}$$

Since $F$ is a continuous function, we know that $\mathcal{Y}$ is an open subset of $\mathbb{R}^n$. Hence, it can be written as the countable union of open boxes in $\mathbb{R}^n$: $$ \mathcal{Y} = \cup_{i=1}^{\infty} B_i $$ where each box $B_i$ has the form: $$ B_i = (a_{1i}, b_{1i}) \times (a_{2i}, b_{2i}) \times ... \times (a_{ni}, b_{ni}) $$ Then \begin{align} \{x \in \mathcal{X} : F(f_1(x), ..., f_n(x)) > c\} &= \{x \in \mathcal{X} : (f_1(x), ..., f_n(x)) \in \mathcal{Y}\}\\ &= \cup_{i=1}^{\infty} \{x \in \mathcal{X} : (f_1(x), ..., f_n(x)) \in B_i\}\\ &= \cup_{i=1}^{\infty} \underbrace{\cap_{j=1}^n \underbrace{\{x \in \mathcal{X} : f_j(x) \in (a_{ji}, b_{ji})\}}_{\mbox{*in $\Sigma$}}}_{\mbox{**in $\Sigma$}} \end{align} This is the countable union of sets in $\Sigma$ and hence is in $\Sigma$. $\Box$

*Note that for each $j \in \{1, ..., n\}$ and each open interval $(a,b)$ we have $$\{x \in \mathcal{X} : f_j(x) \in (a,b)\} \in \Sigma $$ because $$ \{x \in \mathcal{X} : f_j(x) \in (a,b)\} = \cup_{k=1}^{\infty} \left(\{x : f_j(x)>a\} \cap \{x : f_j(x)\leq b-1/k\}\right)$$ and this is a countable union of sets in $\Sigma$, so it is in $\Sigma$.

**Note that we have used the fact that a finite intersection of sets in $\Sigma$ is also in $\Sigma$.

Michael
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  • Defining $h(x) = F(f_1(x), ..., f_n(x))$, the property ${x \in \mathcal{X}:h(x)>c} \in \Sigma$ for all $c \in \mathbb{R}$ implies $h^{-1}(Y) \in \Sigma$ for all subsets $Y \subseteq\mathbb{R}$ that are Borel measurable in $\mathbb{R}$. – Michael Sep 20 '18 at 05:28