Clearly, the improper integral of the function $h: [0,\infty)\to\mathbb R$ exists in
$\mathbb R$ if and only if
$$
\lim_{M,N\to\infty}\int_M^N h(x)\,dx=0.
$$
We shall need the following result:
Lemma. If $F,G :[a,b]\to\mathbb R$ functions, with $G$ Riemann integrable and $F$ monotone, then
$$
\Big| \int_a^b F(x)\,G(x)\,dx\,\Big| \le \big(|F(b)|+|F(b)-F(a)|\big) \cdot\max_{x\in[a,b]}\Big|
\int_a^x G(t)\,dt\,\Big|.
$$
Proof. Let $n\in\mathbb N$, and define the partition of $[a,b]$:
$$
\tau_j= a+j\frac{b-a}{n}, \quad j=0,\ldots,n.
$$
Next we define $F_n: [a,b]\to\mathbb R,$ as a step function, to be equal to
$F(a)$, in the interval $[a,\tau_j)$,
$F(\tau_1)$, in the interval $[\tau_1,\tau_2)\ldots$ and
$F(\tau_{n-1})$, in the interval $[\tau_{n-1},b]$.
Then
\begin{align}
\int_a^b F_n(x)\,G(x)\,dx
&=\sum_{k=1}^n\int_{\tau_{k-1}}^{\tau_k}
F_n(x)\,G(x)\,dx \\
&=\sum_{k=1}^n F(\tau_{k-1})\int_{\tau_{k-1}}^{\tau_k}
G(x)\,dx \\ &\cdots \\&=F(b)\int_a^b g(x)\,dx+\sum_{k=1}^{n-1}
\big(F(\tau_{k-1})-F(\tau_k)\big)\int_a^{\tau_k} g(x)\,dx.
\end{align}
and thus
$$
\Big| \int_a^b F_n(x)\,G(x)\,dx\,\Big| \le \big(|F(b)|+|F(b)-F(a)|\big) \cdot\max_{x\in[a,b]}\Big|
\int_a^x G(t)\,dt\,\Big|, \quad\text{for all}\,\,n\in\mathbb N,
$$
and the proof of the lemma follows from the fact that
$$
\Big| \int_a^b \big(F_n(x)\,G(x)-F(x)\,G(x)\big)\,dx\,\Big|
\le \sup_{x\in [a,b]} |G(x)|\cdot
\Big| \int_a^b \big(F_n(x)-F(x)\big)\,dx\,\Big|,
$$
and clearly $\int_a^b F_n(x)\,dx$ is the lower Riemann sum (or upper sum) corresponding to Riemann integral $\int_a^b F(x)\,dx$, and hence
$$
\Big| \int_a^b \big(F_n(x)-F(x)\big)\,dx\,\Big|\to 0\quad\text{as}\quad n\to\infty. \tag*{$\Box$}
$$
Assume now that $|f_n(x)|, |f(x)|\le K$.
First. For every $n$ the improper integral of the function $f_ng$ exists in $\mathbb R$, as due to the lemma
\begin{align}
\Big|\int_M^N f_n(x)\,g(x)\,dx\,\Big|&\le \big(|f_n(N)|+ |f_n(M)-f_n(N)|\big)\cdot\max_{x\in[M,N]}\Big|
\int_M^N g(t)\,dt\,\Big| \\ &\le 3K \cdot\max_{x\in[M,N]}\Big|
\int_M^N g(t)\,dt\,\Big|.
\end{align}
Clearly, the right hand side tends to zero, as $M,N\to\infty$, and hence
$\lim_{M,N\to\infty}\Big|\int_M^N f_n(x)\,g(x)\,dx\,\Big|=0$, which means that the improper integral of the function $f_ng$ exists in $\mathbb R$.
Second. Let $\varepsilon>0$ and $M>0$, such that
$\sup_{x\ge M}\Big|\int_M^x g(t)\,dx\,\Big|<\frac{\varepsilon}{12K+1}$, and $n_0\in\mathbb N$, such that
$$
|f_n(x)g(x)-f(x)g(x)|<\frac{\varepsilon}{2M},\quad\text{for all}\,\,x\in[0,M]\,\,\text{and}\,\,\, n\ge n_0.
$$
Then
\begin{align}
&\Big|\int_0^\infty f_n(x)\,g(x)\,dx-\int_0^\infty f(x)\,g(x)\,dx\,\Big| \\
&\le
\Big|\int_0^M f_n(x)\,g(x)\,dx-\int_0^M f(x)\,g(x)\,dx\,\Big|+
\Big|\int_M^\infty f_n(x)\,g(x)\,dx-\int_M^\infty f(x)\,g(x)\,dx\,\Big| \\
&\le\int_0^M \big|f_n(x)\,g(x)-f(x)\,g(x)\big|\,dx+
\Big|\int_M^\infty f_n(x)\,g(x)\,dx\Big|+\Big|\int_M^\infty f(x)\,g(x)\,dx\,\Big| \\
&\le M\cdot\frac{\varepsilon}{2M}+2\cdot 3K\cdot \frac{\varepsilon}{12K+1}<\varepsilon,
\end{align}
as
$$
\Big|\int_M^\infty f(x)\,g(x)\,dx\Big|\le
\big(\sup_{x\ge M}|f(x)|+ \sup_{x\ge M} |f(M)-f(x)| \big)\cdot
\sup_{x\ge M}\Big|\int_M^x g(t)\,dx\,\Big|\le 3K\cdot\frac{\varepsilon}{12K+1}.
$$
