$\lim_{t\rightarrow 0}\int_{-\infty}^{\infty}g(x)(f(x)-f(x+t))dx=0$ Let $f$ be integrable over $\mathbb{R}$. Assume $g$ is a bounded measurable function on $\mathbb{R}$. Show that
$$\lim_{t\rightarrow 0}\int_{-\infty}^{\infty}g(x)(f(x)-f(x+t))dx=0.$$
Proof idea: Assume $g$ is bounded measurable function, then there is some $M$ such that $|g(x)|<M$. 
Since $f$ is integrable, $f$ is finite almost everywhere so $f$ is bounded almost everywhere by $h(x)=B$. Since $f(x+t)$ converges to $f(x)$ as $t\rightarrow 0$, by Lebesgue Dominated Convergence theorem, 
$$\lim_{t\rightarrow 0}\int_{\mathbb{R}}|f(x)-f(x+t)|dx=0.$$
Thus, 
$$\left|\lim_{t\rightarrow 0}\int_{-\infty}^{\infty}g(x)(f(x)-f(x+t))dx\right|\leq M\lim_{t\rightarrow 0}\int_{\mathbb{R}}|f(x)-f(x+t)|dx=0.$$
Is the above correct?
 A: As you said there exists $M>0$ such that $|g(x)|\leq M$ for all $x\in\mathbb{R}$. Now let $\epsilon>0$. Since continuous functions with compact support are dense in $L(\mathbb R)$, there exists a continuous function $h\in L(\mathbb R)$ with compact support such that:
\begin{align}
\int_\mathbb{R} |f(x)-h(x)|\,dx <\frac{\epsilon }{3M}
\end{align}
Then:
\begin{align}
\bigg|\int_\mathbb{R}g(x)(f(x+t)-f(x))\,dx \bigg | &\leq \int_\mathbb{R}|g(x)||f(x+t)-f(x)|\,dx\\ &\leq M\int_\mathbb{R}|f(x+t)-f(x)+h(x+t)-h(x+t)|\,dx\\
 &\leq M\int_\mathbb{R}|f(x+t)-h(x+t)|\, dx+M\int_\mathbb{R}|f(x)-h(x+t)|\,dx\\
\end{align}
Now use the DCT as you did to the integral in the RHS to get:
\begin{align}
\lim_{t\to 0} \int_\mathbb{R}|f(x)-h(x+t)|\,dx = \int_\mathbb{R}|f(x)-h(x)|\,dx
\end{align}
Because $h$ is continuous. Hence:
\begin{align}
\lim_{t\to 0} \bigg|\int_\mathbb{R}g(x)(f(x+t)-f(x))\,dx \bigg | 
\leq M\frac{\epsilon}{3M} + M\frac{\epsilon}{3M} < \epsilon 
\end{align}
Since $\epsilon>0$ was arbitrary the result follows.
