Is $\int_{0}^{1}|f_n(x)g_n(x)-f(x)g(x)|\,dx \to 0$ as $ n \to \infty$? True/False Is the following statement true/false  ?

Let $\{f_n\}_{n=1}^{\infty}$ and $f$ be integrable functions on $[0,1]$  such that
$$ \lim_{n \to \infty} \int_{0}^{1}|f_n(x) -f(x)| \, dx = 0 $$
If  $\{g_n\}_{n=1}^{\infty}$ is a uniformly bounded sequence of continuous functions converging pointwise to a function $g$, then
$$ \int_{0}^{1} |f_n(x)g_n(x)-f(x)g(x)| \, dx \to 0 \qquad\text{as}\quad  n \to \infty $$

My attempt :  I think this  statement is false. Take $f_n(x) = f(x) = 1$ for all $x \in [0,1]$. Let $g_n(x)=n \sin n x$ if $x \in [0,\pi/n]$ and zero elsewhere. But here $\int_{0}^{1} g_n(x) \,dx = 2 \neq 0$.
 A: Commentary:
I have the following question about integrable functions.  Suppose that $\{f_n\}$ is a sequence of integrable $\dots$
Hold on there!  If you are deep inside a class lecture or an analysis textbook then "integrable" has a well-defined meaning.  But out of context this has no meaning.  All too often students post such questions and it is not clear at what level the question is being posed.
For the 18th century it would have meant the integral of Leibnitz and Newton.  For a calculus student it invariably means the Riemann integral.  For an advanced student it means, no doubt the Lebesgue integral.  But, even so, the meaning might be the improper Riemann integral or the improper Lebesgue integral.  For a research mathematician there are dozens of different interpretations (Denjoy-Perron, Denjoy-Khintchine, approximate Perron, Cesàro-Perron, etc.).
So ...
PART I:  If the OP intends the Lebesgue integral then there is an excellent answer already posted by @KaviRamaMurthy  and already accepted.
PART II: If the OP intends the Riemann integral then we are working with a narrow interpretation of "integrable function" and a much narrower toolkit to work with it.
In fact, for this question:

*

*The statement in the question posed is false.   There is the possibility that the function $g$ stated there is not Riemann integrable.  It must be bounded, of course, but it need not be Riemann integrable nonetheless.  For a counterexample select any differentiable Lipschitz function $G$   whose derivative $g=G'$ is not Riemann integrable on $[0,1]$.  Volterra gave one in the 1870s, famous enough that a search on Wikipedia or StackExchange will easily find it.  In that case, the sequence$$g_n(x) = n\left[G\left(x+\frac1n\right) -G(x)\right] \to G'(x)=g(x)$$
is uniformly bounded and converges pointwise to a function that is not at all Riemann integrable on $[0,1]$.


*If you add to the question the assumption that $g$ is Riemann integrable then the statement is true.   One way to see this is that the other answer for the Lebesgue integral supplies this automatically since the Lebesgue integral includes the Riemann integral.   Alternatively you can learn the Arzelà-Osgood bounded convergence theorem for the Riemann integral (dates back to the late 19th century also).   In that case observe that$$ |f(x)[g_n(x)-g(x)]|  \to 0 \qquad\text{as}\quad  n \to \infty$$ is a uniformly bounded sequence of Riemann integrable functions.  As $f$ and $g$ are Riemann integrable they are bounded, and the sequence $\{g_n\}$ is also uniformly bounded by hypothesis.  Hence, by the ancient 1880s bounded convergence theorem $$\int_{0}^{1} |f(x)[g_n(x)-g(x)]| \, dx \to 0 \qquad\text{as}\quad  n \to \infty.\tag{1}$$  Now use the clever device from the other answer of @KaviRamaMurthy  $$\int_{0}^{1}|f_n(x)g_n(x)-f(x)g(x)|\,dx$$ $$\leq \int_{0}^{1}|g_n(x)||f_n(x)-f(x)|\,dx+\int_{0}^{1}|f(x)||g_n(x)-g(x)|\,dx.$$   The second integral on the right is already handled in (1) and the first integral on the right is easy since $$\int_{0}^{1}|g_n(x)||f_n(x)-f(x)|\,dx \leq M \int_{0}^{1} |f_n(x)-f(x)|\,dx \to 0$$
where $M$ is a uniform bound for the sequence $\{g_n\}$.

Selected Bibliography of the classical Bounded Convergence Theorem:
This is the theorem (for the Riemann integral) that anticipated Lebesgue's more famous version for his integral. It is accessible to any analysis student learning the Riemann integral and needs no measure theory.

*

*C. Arzelà, Sulla integrazione per serie, Atti Acc. Lincei Rend., Rome (4) 1 (1885), 532–537 and 596–599.


*P. S. Bullen and R. Vborny, Arzelà’s dominated convergence theorem for the Riemann integral, Boll. Un.
Mat. Ital. A (7) 10 (1996), no. 2, 347–353.


*F. Cunningham Jr. Taking limits under the integral sign, Mathematics Magazine, vol. 40, 1967, 179–186.


*W. F. Eberlein, Notes on integration. I. The underlying convergence theorem. Comm. Pure Appl. Math.
10 (1957), 357–360.


*H. Kestelman, Classroom Notes: Riemann integration of limit functions, Amer. Math. Monthly 77 (1970),
no. 2, 182–187.


*H. A. Lauwerier, An elementary proof of the Arzelà-Osgood-Lebesgue theorem, (Dutch) Simon Stevin
26, (1949) 177–179.


*Jonathan W. Lewin, A truly elementary approach to the bounded convergence theorem, Amer. Math.
Monthly, Vol. 93, No. 5 (1986), 395–397.


*W. A. J. Luxemburg, Arzelà’s dominated convergence theorem for the Riemann Integral, Amer. Math.
Monthly, Vol. 78, No. 9 (1971), 970–979.


*William F. Osgood, Non-uniform convergence and the integration of series term by term, American
Journal of Mathematics, 19 (1897), 155–190.


*B. Thomson (2020) The Bounded Convergence Theorem, The American Mathematical Monthly, 127:6, 483-503
A: The statement is true.
$$\int_{0}^{1}|f_n(x)g_n(x)-f(x)g(x)|dx\leq \int_{0}^{1}|g_n(x)||f_n(x)-f(x)|dx\\~~~~~~~~ ~~~~~~~ ~~~~~~~ ~~~~~~~ ~~~~~~~ ~~~~~~~ ~~~~~+\int_{0}^{1}|f(x)||g_n(x)-g(x)|dx.$$
The first term clearly tends to $0$. For the second term split the integral into integral over $\{x: |g_n(x)-g(x)| <\epsilon\}$ and its complement. The first of these parts tends to $0$ as $\epsilon \to 0$. For the second one use the following facts:

*

*Measure of $\{x: |g_n(x)-g(x)| \geq \epsilon\}$ tends to $0$ since convergence a..e implies convergence in measure.


*$\int_A |f(x)|dx \to 0$ as the measure of $A$ tends to $0$.
