# repeated integration of lebesgue integrable function

I am currently working with Wheeden Zygmund's Measure and Integral. I got stuck on one of the exercises (Ch. 6, Ex. 13).

Let $f \in L(-\infty,\infty)$ (Lebesgue integrable), and let $h > 0$ be fixed. Prove that $$\int_{-\infty}^\infty (\frac{1}{2h} \int_{x-h}^{x+h} f(y) \, dy) \, dx = \int_{-\infty}^\infty f(x) \, dx$$

I believe it suffices to prove that $\frac{1}{2h} \int_{x-h}^{x+h} f(y) \, dy = f(x)$ using the mean value theorem, but I am not entirely sure. Can anyone give me a hint on this problem? Thanks in advance!

• Hint: Fubini's theorem. Commented Mar 1, 2017 at 13:34
• thanks for the hint. could you possibly show how i can apply fubini's thm here? Commented Mar 2, 2017 at 7:36

Just note by Fubini's theorem that \begin{align*} \frac{1}{2h}\int_{-\infty}^\infty \int_{x-h}^{x+h} f(y) dy dx &= \frac{1}{2h}\int_{-\infty}^\infty \int_{-\infty}^{\infty} 1_{(x-h, x+h)}(y) \cdot f(y) dy dx \\ &= \frac{1}{2h}\int_{-\infty}^\infty f(y) \cdot \int_{-\infty}^{\infty} 1_{(y-h, y+h)}(x) dx dy = \int_{-\infty}^\infty f(y) dy, \end{align*} since $1_{(x-h,x+h)}(y) = 1_{(y-h, y+h)}(x)$.