Prove that: $\lim_{t\to\infty}\frac{1}{t_n}\int_{1}^{t}{f(s)g(s)}\,ds=0$ Maybe someone can help me out with the following question:
Let $f \in L(E)$ (Lebesgue integrable in E) and g $\in M(E)$ (measurable in $E$). If the function $h(t)=g(t)/t$ is bounded, prove that:
$$\lim_{t\to\infty}\frac{1}{t}\int_{1}^{t}{f(s)g(s)}\,ds=0$$
I´ve advanced until:
I´ve taken a sequence {$t_n$} / {$t_n$}$\to\infty$. As
$\lim_{t_n\to\infty}\frac{1}{t_n}\int_{1}^{t_n}{f(s)g(s)}\,ds=0$$\iff$$\lim_{t\to\infty}\frac{1}{t}\int_{1}^{t}{f(s)g(s)}\,ds=0$.
Therefore:
$$\begin{align*}
\lim_{t_n\to\infty} \frac{1}{t_n} \int_{1}^{t_n}{f(s)g(s)}\,ds &= \lim_{t_n\to\infty} \frac{1}{t_n} \int_{1}^{t_n}\frac{f(s)g(s)s}{s}\,ds\\
&\leq \lim_{t_n\to\infty}\frac{M}{t_n}\int_{1}^{t_n}f(s)s=\lim_{t_n\to\infty}\frac{M}
{t_n}\int_{1}^{\infty}\chi_{[1,t]}f(s)s
\end{align*}$$
But I can´t continue, I might need to use the Dominated convergence theorem but I don´t know how to use it. Can someone help me?
 A: Define $F(t)=\frac{1}{t}\int_{1}^{t}f(x)g(x)dx$. To prove that $\lim_{t\rightarrow\infty}F(t)=0$,
it suffices that for each sequence $(t_{n})$ in $[1,\infty)$, if
$t_{n}\rightarrow\infty$, then $F(t_{n})\rightarrow0$ (a result
due to Heine).
Choose $M>0$ such that $\left|\frac{g(t)}{t}\right|\leq M$ for each
$t\in[1,\infty)$. Let $(t_{n})$ be an arbitrary sequence in $[1,\infty)$
such that $t_{n}\rightarrow\infty$. Observe that for each $n$ and
each $x\in[1,\infty)$, we have
\begin{eqnarray*}
 &  & \left|\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)\right|\\
 & \leq & \left|\frac{x}{t_{n}}1_{[1,t_{n}]}(x)f(x)\cdot\frac{g(x)}{x}\right|\\
 & \leq & M\left|\frac{x}{t_{n}}1_{[1,t_{n}]}(x)f(x)\right|\\
 & \leq & M\left|1_{[1,t_{n}]}(x)f(x)\right|\\
 & \leq & M|f(x)|.
\end{eqnarray*}
Moreover, for each fixed $x\in[1,\infty)$, $\lim_{n\rightarrow\infty}\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)=0$
because $\frac{1}{t_{n}}\rightarrow0$. By Lebesgue Dominated Convergence
Theorem, we have that $\lim_{n\rightarrow\infty}\int\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)dx=0$.
Note that $\int\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)dx=\frac{1}{t_{n}}\int_{1}^{t_{n}}f(x)g(x)dx=F(t_{n})$.
That is, $F(t_{n})\rightarrow0$.
