Application of Fubini: $\int^{\infty}_{-\infty}\int^{x+h}_{x-h}=\int^{\infty}_{-\infty}\int^{y+h}_{y-h}$ 
Let $f$ be Lebesgue integrable on $(-\infty,\infty)$, and $h>0$ be fixed. Show that
$$\int^{\infty}_{-\infty}\left(\frac{1}{2h}\int^{x+h}_{x-h}f(y)\,dy\right)\,dx=\int^{\infty}_{-\infty}f(x)\,dx\text{.}$$

Fubini's Theorem

Suppose $f(x,y)$ is integrable on $\mathbb{R}^{d_1}\times\mathbb{R}^{d_2}$. Then
$$\int_{\mathbb{R}^{d_2}}\left(\int_{\mathbb{R}^{d_1}}f(x,y)\,dx\right)\,dy=\int_{\mathbb{R}^{d_1}}\left(\int_{\mathbb{R}^{d_2}}f(x,y)\,dy\right)\,dx=\int_{\mathbb{R}^d}f\text{.}$$

The answer is

$$\begin{align*}
\int^{\infty}_{-\infty}\left(\frac{1}{2h}\int^{x+h}_{x-h}f(y)\,dy\right)\,dx&=\frac{1}{2h}\int^{\infty}_{-\infty}\left(\int^{y+h}_{y-h}f(y)\,dx\right)\,dy\\
&=\int^{\infty}_{-\infty}f(y)\,dy\text{.}
\end{align*}$$

How to justify the first equality? I know that
$$\int^{\infty}_{-\infty}\left(\frac{1}{2h}\int^{x+h}_{x-h}f(y)\,dy\right)\,dx=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}(f\chi_{[x-h,x+h]})(y)dydx\text{.}$$
 A: Since $x-h\leq y\leq x+h$ if and only if $y-h\leq x\leq y+h$,
$$\begin{align}
\int^\infty_{-\infty}\left(\int^{x+h}_{x-h}f(y)\,dy\right)dx&=\int^{\infty}_{-\infty}\left(\int^{\infty}_{-\infty}f(y)\chi_{[x-h,x+h]}(y)\,dy\right)dx\\
&=\int^{\infty}_{-\infty}\left(\int^\infty_{-\infty}f(y)\chi_{[x-h,x+h]}(y)\,dx\right)dy\\
&=\int^\infty_{-\infty}\left(\int^{y+h}_{y-h}f(y)\chi_{[x-h,x+h]}(y)\,dx\right)dy\\
&=\int^{\infty}_{-\infty}\left(\int^{y+h}_{y-h}f(y)\,dx\right)dy\\
&=\int^\infty_{-\infty}\left(2hf(y)\right)dy\\
&=2h\int^{\infty}_{-\infty}f(y)\,dy\text{,}
\end{align}$$
where the second equality follows from Fubini's theorem. To justify its use, we have to prove the integrability of $F(x,y)=f(y)\chi_{[x-h,x+h]}(y)$. By Tonelli's theorem,
$$\begin{align}
\int |F|&=\int^\infty_{-\infty}\left(\int^\infty_{-\infty}|f(y)||\chi_{[x-h,x+h]}(y)\,dx\right)dy\\
&=\int^\infty_{-\infty}\left(\int^{y+h}_{y-h}|f(y)|\,dx\right)dy\\
&=2h\int^{\infty}_{-\infty}|f(y)|\,dy<\infty\text{.}
\end{align}$$
This completes the proof.
A: The constant $\frac{1}{2h}$ is negligible, and the only thing you need to prove is $$f(x)\chi_{[x-h, x+h]}(y)\in L^1(\mathbb{R}\times\mathbb{R}).$$ This can be done by Tonelli's theorem(that the Fubini's theorem always holds for non-negative measurable $f(x, y)$). We have $$\int_{\mathbb{R}^2}|f(x)\chi_{[x-h, x+h]}(y)| = \int_{\mathbb{R}}\int_{\mathbb{R}}|f(x)|\cdot\chi_{[x-h, x+h]}(y)dxdy = \int_{\mathbb{R}}|f(x)|dx\cdot\int_{\mathbb{R}}\chi_{[x-h, x+h]}(y)dy < \infty$$ by the condition that $f$ is Lebesgue integrable.
