# Continuity and lebesgue integrability of integral function, proof verification

Suppose that $$f\in L^1(\mathbb R, \mathcal B_{\mathbb R} ,m)$$ where $$m$$ is the standard lebesgue measure. For fixed $$h$$, let us define: $$\phi(x)= \frac{1}{2h} \int_{x-h}^{x+h} f(t) dt$$ Show that $$\phi$$ is measureable and that $$\|\phi \|_1 \leq \|f\|_1$$.

For the first part, I have tried the following:

We will show that $$\phi$$ is continuous on $$\mathbb R$$. For this, let $$x_n \to x$$ be an arbitrary sequence in $$\mathbb R$$. We define, $$E_n = [x_n -h , x_n +h]$$ and $$E= [x-h,x+h]$$. Note that $$f\chi_{E_n} \to f\chi_{E}$$ pointwise almost since for any $$y \in \mathbb R$$ (I feel like this is intuitively obvious, but I'm having a little trouble formulating the details). Next, $$|f \chi_{E_n}| \leq |f|$$, and since $$f\chi_{E_n}, f\chi_E \in L^1$$, we have by the DCT that $$\int f\chi_{E_n} \to \int f\chi_E \text{ as } n\to \infty$$ which implies that for $$n$$ sufficiently large, $$\frac{1}{2|h|} \left| \int_{E_n} f - \int_E f \right| \leq \epsilon$$ i.e. $$\phi(x_n) \to \phi(x)$$.

I am having trouble with the second part. Any suggestions?

For the convergence $$\chi_{E_n}\rightarrow\chi_E$$, consider points $$t$$ in $$\mathbb{R}$$ different from from $$x+h$$ and $$x-h$$; then there is a positive $$\delta$$ for which $$(t-\delta,t+\delta)$$ is either contained in it $$(x-h,x+h)$$ or its compliment. So, for sufficiently large $$n$$, we have that $$\vert x_n-x\vert<\delta$$ and then it should hold that $$\chi_{E_n}(t)=\chi_E(t)$$.
For the second part, $$\Vert\phi\Vert_1=\int_\mathbb{R}\vert\phi(t)\vert\ dt=\int_{\mathbb{R}}\left\vert\frac{1}{2h}\int_{t-h}^{t+h}f(x)\ dx\right\vert dt\le\frac{1}{2\vert h\vert}\int_\mathbb{R}\int_{t-h}^{t+h}\vert f(x)\vert\ dxdt$$ and by Fubini's theorem, that equals $$\frac{1}{2\vert h\vert}\int_\mathbb{R}\int_{x-h}^{x+h}\vert f(x)\vert\ dtdx=\frac{1}{2\vert h\vert}\int_\mathbb{R}2\vert h\vert\vert f(x)\vert\ dx=\int_\mathbb{R}\vert f\vert\ dx=\Vert f\Vert_1.$$ Notice that the region of integration is $$\{(x,t)\ \vert\ t-h\le x\le t+h\}$$ which is equal to $$\{(x,t)\ \vert\ x-h\le t\le x+h\}$$, so that is why Fubini works like that (plus details of convergence, but those are standard).For every real $$x$$, we have that $$\int_{x-h}^{x+h}\vert f(x)\vert\ dt=2\vert h\vert\vert f(x)\vert$$ simply becuase $$f(x)$$ is constant with respect to $$t$$.
I will asume that $$h>0$$. The case $$h<0$$ is left to you.
$$2h\|\phi_1\|=\int_{-\infty}^{\infty} |\int_{x-h}^{x+h} f(t) dt| dx$$ $$\le \int_{-\infty}^{\infty} \int_{t-h}^{t+h} |f(t)| dx dt$$ $$=\int_{-\infty}^{\infty}2h |f(t)| dt$$ $$=2h||f\|_1 .$$
I have used the fact that $$x-h is equivalent to $$t-h. Interchange of integrals is justified by Tonelli's Theorem.