$\phi_h(x)=\int_{x+h}^{x+h}f(t)dm(t)$ where $f \in \mathcal{L}^1(\mathbb{R}, \mathcal{B}(\mathbb{R}), m)$

Question

Let $f \in \mathcal{L}^1(\mathbb{R}, \mathcal{B}(\mathbb{R}), m)$. Fix $h>0$. Define $\phi_h(x)=\int_{x+h}^{x+h}f(t)dm(t)$. Prove that
(1) $\phi_h$ is $\mathcal{B}(\mathbb{R})-$measurable.
(2) $\phi_h \in \mathcal{L}^1(\mathbb{R}, \mathcal{B}(\mathbb{R}), m)$.
(3) $\int_{\mathbb{R}} \vert \phi_h \vert dm \leq 2h \cdot \int_{\mathbb{R}}\vert f \vert dm$.

Can anyone give me some hints? For (1), I think I can't check if the inverse image of $\phi_h$ on each open set is open as usual cause $f$ is general.

Answer 1

For $x_{n}\rightarrow x$, $x_{n}>x$, then $\chi_{[x_{n}-h,x_{n}+h]}\rightarrow\chi_{[x-h,x+h]}$ pointwise. Now $|\chi_{[x_{n}-h,x_{n}+h]}(t)f(t)|\leq |f(t)|$, so Lebesgue Dominated Convergence Theorem implies that $\phi_{h}(x_{n})\rightarrow\phi_{h}(x)$. Similarly for $x_{n}<x$, $x_{n}\rightarrow x$, so $\phi_{h}$ is actually continuous.

\begin{align*} \int_{{\bf{R}}}|\phi_{h}(x)|dm(x)&\leq\int_{{\bf{R}}}\int_{{\bf{R}}}\chi_{[x-h,x+h]}(t)|f(t)|dm(t)dm(x)\\ &=\int_{{\bf{R}}}\int_{{\bf{R}}}\chi_{[x-h,x+h]}(t)|f(t)|dm(x)dm(t)\\ &=\int_{{\bf{R}}}\int_{t-h}^{t+h}dm(x)|f(t)|dt\\ &=\int_{\bf{R}}2h|f(t)|dt. \end{align*}