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Problem Let $[a,b] \subset \mathbb R$ and $f: \mathbb R \to \mathbb R_{\geq 0}$ such that $f=0$ on $[a,b]^c$. Given $h >0$, we define $g: \mathbb R \to \mathbb R_{\geq 0}$ given by $$g(x)=\dfrac{1}{2h}\int_{x-h}^{x+h} f(t)dt$$

Prove that $g$ is measurable and that $$\int_a^b g(x)dx \leq \int_a^b f(x)dx$$

I thought of defining the function $$\phi(x,t)=\dfrac{1}{2h}\chi_{[x-h,x+h]}(t)f(t),$$ so for each $x$, $\phi(x,t)$ is a measurable function of variable $t$ and $g(x)=\int_{\mathbb R} \phi(x,t)dt$

But I have no idea how to deduce from here that $g(x)$ is measurable and how to prove the inequality of the exercise. My idea was to construct a measurable non negative function $T(x,t)$, then apply Fubini so as to be able to affirm that $g(x)=\int_{\mathbb R} T(x,t)dt$, if I could do that, it follows that $g$ is measurable. As for the inequality part, I am still lost.

Any help from hints and suggestions would be greatly appreciated.

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  • $\begingroup$ What do you mean by $[a, b]^c$? The set is already closed. $\endgroup$ Oct 3, 2015 at 19:13
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    $\begingroup$ @PaulSinclair: it is the standard notation for the complement of the set $\endgroup$
    – Giovanni
    Oct 3, 2015 at 19:13
  • $\begingroup$ Be careful with assuming notations are "standard". I've seen numerous notations of the complement of the set. And I've far more often seen this notation used for the closure of a set. $\endgroup$ Oct 3, 2015 at 19:26
  • $\begingroup$ Thanks for the good advise Paul, I'll be more careful in the future. $\endgroup$
    – Giovanni
    Oct 3, 2015 at 19:34

1 Answer 1

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Because $f$ is nonnegative, you are free to use Tonelli's Theorem. This is the tool needed to make your approach work.

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  • $\begingroup$ I've already tried to come up with a function to apply Tonelli's (the function $\phi$ that I define in my post) but I am not sure if it's measurable. $\endgroup$
    – user16924
    Oct 5, 2015 at 15:48

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