Let $f:\mathbb{R} \rightarrow (0, \infty)$ be a Borel measurable function and let $E$ be a Borel measurable subset of $\mathbb{R}$ such that $\lambda (E) > 0$. Define $F(t) = \int_{E}f(t + x) d \lambda (x)$, $t \in \mathbb{R}$.
Prove that $F$ is a Borel measurable. Prove further that if $F \in L^{1}(\mathbb{R})$, then $f \in L^{1}(\mathbb{R})$ and $\lambda (E) < \infty$.
We can write $f(s) = \sum_{i=1}^{n}a_{i}\chi_{F_{i}}(s)$, where $F_{i}$ are pairwise disjoint Borel sets and $a_{i}$ are positive real numbers. Then $F(t) = \sum_{i=1}^{n}a_{i}\lambda (E \cap (F_{i} - t))$. I think this is the point where I need to invoke $\lambda (E) > 0$.
Let $\phi(t,x)=1_E(x)f(t-x)$. Then $\phi\in L^+(\mathbb{R}\times\mathbb{R})$ (it is nonnegative and measurable w.r.t the product $\sigma$-algebra) and by (Fubini-)Tonelli theorem $F(t)=\int \phi_td\lambda$ is measurable (if $\lambda(E)=0$, then $F\equiv 0$ so that it's trivially measurable).
As for the second part, by (Fubini-)Tonelli again and translation invariance of the Lebesgue integral, $$ \int F=\int_{\mathbb{R}}\int_Ef(t+x)d\lambda(x)d\lambda(t)=\int_E\int_{\mathbb{R}}f(t+x)d\lambda(t)d\lambda(x)\\ =\int_E\int_{\mathbb{R}}f(t)d\lambda(t)d\lambda(x)=\int f\times \lambda(E)< \infty. $$