Let $f\in L^{1}(\mathbb{R})$ and $\forall x\in\mathbb{R},$ let $F(x)=\displaystyle\int_x^{x+1} f(t)\ dt$. Prove that $F\in L^{1}(\mathbb{R})$. Hint: Use Tonelli.
Attempt: I indeed used Tonelli's theorem. Upon switching the order of integration, I was able to reach the point where we have $\displaystyle\int_{\mathbb{R}} |F(x)|\ dx\leq\displaystyle\lim_{a\to\infty} I_a$, where $$I_a=\displaystyle\int_{-a}^{-a+1}|f(t)|(t+a)\ dt+\displaystyle\int_{-a+1}^{a}|f(t)|\ dt+\displaystyle\int_{a}^{a+1}|f(t)|(a-t+1)\ dt.$$ My only stumbling block is the part where I have to compute $\displaystyle\lim_{a\to\infty}\left(a\displaystyle\int_{-a}^{-a+1}|f(t)|\ dt\right)$ which gives an indeterminate form. Thus this converge to $0$? Thanks for any help.
Do not split integrals. $\int_{\mathbb R} |F((x)|dx\leq \int_{\mathbb R} \int_x^{x+1} |f(t)|dtdx=\int_{\mathbb R} \int_{t-1}^{t} |f(t)|dxdt=\int_{\mathbb R} |f(t)|dt<\infty$ because $\int_{t-1}^{t}dx=1$.