Extending the integral of a bounded measurable function from a measurable set $E$ to the whole of $\mathbb{R}$ 
Let $g(x)$ be a bounded measurable function such that $$\lim_{n \rightarrow \infty} \int_E g(nx)dx = 0$$ for any measurable set $E$ with finite measure. Given $f \in L^1(\mathbb{R})$, do we have $$\lim _{n \rightarrow \infty} \int _{\mathbb{R}} f(x)g(nx)dx = 0?$$

My approach is to partition $\mathbb{R}$ into a countable collection of intervals, say, $E_i=[i, i+1]$, and we consider the integral $$\int_{E_i} f(x)g(nx) dx.$$
Since $g$ is bounded, we can find $M_n \in \mathbb{R}$ such that $$-M_nf(x) \leq f(x)g(nx) \leq M_nf(x)$$ and so the integral will be bounded above by $$M_n\int_{E_i}f(x)dx,$$
which is finite. By hypothesis, we have that $$\lim _{n \rightarrow \infty} \int_{E_i} g(nx)dx = 0$$ and so the sequence of upper bounds $\{M_n\}$ converges to zero as well. Hence $$\lim _{n \rightarrow \infty} \int _{E_i} f(x)g(nx)dx = 0.$$
Could we then conclude that $$\lim _{n \rightarrow \infty} \int _{\mathbb{R}} f(x)g(nx)dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^\infty \int _{E_i} f(x)g(nx)dx = \sum_{i=1} ^\infty \lim_{n \rightarrow \infty}\int _{E_i} f(x)g(nx)dx = 0?$$
I can't properly justify why I could switch the summation and limit and I'm not very good at manipulating the arbitrary integral of a product of functions, any advice would be appreciated.
 A: Try to prove this lemma:
Let $a_{m,n}$ be a real-valued sequence on $m\in\mathbb{Z}$, $n\in\mathbb{Z}^+$.
Suppose that $|a_{m,n}|\leq r_m$ for all $m$ and $n$,
that $\displaystyle \sum_{m\in\mathbb{Z}}r_m$ converges,
and that for each $m$ we have $\displaystyle\lim_{n\to\infty}a_{m,n}=s_m\in\mathbb{R}$.
Then obviously $\displaystyle\sum_{m\in\mathbb{Z}}s_m$ converges,
and for each $n$ we find that $\displaystyle\sum_{m\in\mathbb{Z}}a_{m,n}$ converges,
and also $\displaystyle\lim_{n\to\infty}\sum_{m\in\mathbb{Z}}a_{m,n}$ exists and is equal to $\displaystyle\sum_{m\in\mathbb{Z}}s_m$.
$\\$
Because $f$ is in $L_1(\mathbb{R})\supset L_1[j,j+1]$
and $|g(x)|<M$ for all $x\in\mathbb{R}$,
we have
$\displaystyle\bigg|\int_{[j,j+1]}f(x)g(nx)\,\text{d}x\bigg|$
$\leq M\displaystyle\int_{[j,j+1]}|f(x)|\,\text{d}x$.
Also, from $f\in L_1(\mathbb{R})$ itself,
it follows that $\displaystyle\sum_{j=-\infty}^\infty \int_{[j,j+1]}|f(x)|\,\text{d}x$ converges.
Obviously then $\displaystyle\sum_{j=-\infty}^\infty M\int_{[j,j+1]}|f(x)|\,\text{d}x$ converges.
Therefore,
$\displaystyle\lim_{n\to\infty}\sum_{j=-\infty}^\infty\int_{[j,j+1]}f(x)g(nx)\,\text{d}x$
$=\displaystyle\sum_{j=-\infty}^\infty\lim_{n\to\infty}\int_{[j,j+1]}f(x)g(nx)\,\text{d}x$
$=0$.
