Prove that a intergral over $\mathbb R$ is finite Let $K\in \mathcal L_1(\mathbb R)$ and $f$ be measurable and bounded on $\mathbb R$ such that $\lim_{|x|\to \infty} f(x)=0$. Define $$F(x):= \int _{\mathbb R} K(x-s)f(s)\;ds \qquad (x\in \mathbb R)$$
Prove that $F(x)$ is finite for all $x\in \mathbb R$ and that $\lim_{|x|\to \infty} F(x)=0$. I was told that the proof would probably include the use of the Fubini's theorem or Tonelli's Theorem, but I don't see how they apply here... 
 A: For the first part, since $f$ is bounded:
$$
|F(x)| = \left|\int_{\mathbb R} K(x-s)f(s)\,ds\right| =
\left| \int_{\mathbb R} f(x-s)K(s)\,ds \right| \le M \int_\mathbb R |K(s)|\,ds
$$
where $M = \sup_\mathbb{R} |f|$. (The absolute values are missing in @Ant's solution.)
For the second part,
\begin{align}
\lim_{x \to \pm\infty} F(x) &= 
\lim_{x \to \pm\infty} \int_{\mathbb{R}} K(x-s)f(s)\,ds \\
&= \lim_{x \to \pm\infty} \int_{\mathbb{R}} f(x-s)K(s)\,ds \\
&= \int_{\mathbb{R}} \lim_{x \to \pm\infty} f(x-s)K(s)\,ds = 0
\end{align}
by dominated convergence. (Note that $|f(x-s)K(s)| \le M |K(s)|$ for all $x$, and the right hand side in the inequality is $L^1$.)
A: Note that $f$ is bounded on $\mathbb R$, so set $M = \sup |f(x)|$, then $$F(x) = \int_\mathbb R K(x-s)f(s)ds  = \int_\mathbb Rf(x-s)K(s)ds\le M \int_\mathbb R K(s)ds$$
and since $K \in \mathcal L^1(\mathbb R)$ you are done.
To show that $\lim_{x \to \infty} F(x) = 0$, note that for every $\varepsilon > 0$, exists $N > 0$ such that for every $x > N$, $|f(x)| < \varepsilon$. But this also implies that $$\left|\int_\mathbb R K(x-s)f(s)ds\right| \le \int_\mathbb R \left | K(x-s)f(s)ds\right| =  \int_\mathbb R |K(x-s)| \cdot |f(s)| ds \le \varepsilon \int_\mathbb R |K(x-s)|ds$$
Since $K \in \mathcal L^1(\mathbb R)$, set $L = \int_\mathbb R |K(x-s)|ds \in \mathbb R$. This is the exact definition of $\lim_{x \to \infty} F(x) = 0$ !
That is to say, for every $\varepsilon > 0$ exists $N$ such that for every $x > N$ we have $$|F(x)| = \left|\int_\mathbb R K(x-s)f(s)ds\right| \le \varepsilon L$$
(note that the fact that $\varepsilon$ is multiplied by a constant does not change anything ) 
