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?