Problem. Given $f \in L(\mathbb{R})$, let $$\phi_{h}(x) = \frac{1}{2h}\int_{x-h}^{x+h}f(t)dt,\quad h>0.$$ Prove that $\phi \in L(\mathbb{R})$ e $\displaystyle \int_{\mathbb{R}}|\phi_{h}(x)|dx \leq \Vert f \Vert_{1}.$
My attempt.
\begin{eqnarray*} \int_{\mathbb{R}}|\phi_{h}(x)| & \leq & \frac{1}{2h}\int_{\mathbb{R}}\int_{[x-h,x+h]}|f(t)|dtdx\\ & = & \frac{1}{2h}\underbrace{\int_{\mathbb{R}}\int_{\mathbb{R}}|f(t)|\chi_{[x-h,x+h]}dtdx}_{\text{Fubini-Tonelli}}\\ & = & \frac{1}{2h}\int_{\mathbb{R}}\int_{\mathbb{R}}|f(t)|\chi_{[x-h,x+h]}dxdt\\ & = & \frac{1}{2h}\int_{\mathbb{R}}|f(t)|\underbrace{m([x-h,x+h])}_{m\text{ is the Lebesgue measure}}dt\\ & = & \int_{\mathbb{R}}|f(t)|dt\\ & = & \Vert f \Vert_{1} \end{eqnarray*}
Is this correct?