show that $ f_h \in L^1(R) $ and $ \lim_{h \to 0} f_h(x) = f(x) $ in $ L^1(R) $

Question

If $ f \in L^1(R) $ and set $ f_h(x)= \frac{1}{2h} \int_{x-h}^{x+h}f(t)dt, h>0 $ then show that $ f_h \in L^1(R) $ and $ \lim_{h \to 0} f_h(x) = f(x) $ in $ L^1(R) $

To prove f_h is integrable in R we have to this integral is finite almost everywhere. I just wondering how to do that? looking for some hints

Answer 1

It follows from the triangle inequality and a change of variables that

$$\int_{\mathbb{R}} |f_h(x)| \, dx \leq \frac{1}{2h} \int_{\mathbb{R}} \int_{x-h}^{x+h} |f(y)| \, dy \, dx = \frac{1}{2h} \int_{\mathbb{R}} \int_{-h}^h |f(x+t)| \, dt \, dx.$$

Now use Tonelli's theorem to conclude $\|f_h\|_1 \leq \|f\|_1$.