Let $f_n(x)$ be a sequence of integrable functions such that $\lim_{n\to \infty} \int^a_b |f_n(x)|dx=0$. Prove the following

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If $g$ is an integrable function then

$$\lim_{n\to \infty} \int^a_b g(x)|f_n(x)|dx=0$$

Ideas : Intuitively, it is somewhat obvious, but I don't know how to show it.. essentially, the function we must integrate is the product of two integrable functions, so it is integrable.

Then it suffices to calculate the limit of the infimum of the Upper Riemann Sum (or the limit of the supremum of the Lower Riemann sum), and show that it must equal 0. But I don't know how to show this.. I'll have to use what it means for a sequence to converge along with the fact that $g$ is integrable and $f_n$ is integrable, but I don't know how to bring these pieces together.

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It is about Riemann integrals: Since $g$ is integrable on $[a,b]$, so $M:=\sup_{x\in[a,b]}|g(x)|<\infty$, and hence \begin{align*} \int_{a}^{b}|g(x)|\cdot|f_{n}(x)|dx\leq M\int_{a}^{b}|f_{n}(x)|dx, \end{align*} now use Squeeze Theorem to conclude:

Let $u_{n}=\displaystyle\int_{a}^{b}|g(x)|\cdot|f_{n}(x)|dx$ and $v_{n}=M\displaystyle\int_{a}^{b}|f_{n}(x)|dx$, then $0\leq u_{n}\leq v_{n}$ and $v_{n}\rightarrow 0$. But $0\rightarrow 0$, so $u_{n}\rightarrow 0$.