I was trying to prove the following and had totally no idea how to start.
Let $f$ be integrable over $(-\infty, \infty)$, and let $h> 0$ be fixed. Prove that $$\int_{-\infty}^{\infty} (\frac{1}{2h} \int_{x-h}^{x+h} f(y) \mathrm{d}y) \mathrm{d}x = \int_{-\infty}^{\infty}f(x) \mathrm{d}x$$
When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.
LHS $=\frac 1 {2h}\int_{-\infty}^{\infty} \int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.