Showing $\int_{a}^b |f_n(x)| dx \to 0$ as $n \to \infty$

Question

I have a problem where I have to show $\int_{a}^b |f_n(x)| dx \to 0$ as $n \to \infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, **$

I am thinking to first try to show that $f_n$ uniformly converges to zero as in for every $\epsilon > 0$, $\exists N$ such that $n \geq N$ $\implies $ $\sup \limits_{x \in [a,b]} |f_n(x)| \leq \epsilon$.

But is uniform convergence to zero of $f_n$ even a necessary condition to show $**$.

In other words, if I fail to show uniform convergence, can I still hope to show $**$. If yes, what are some of the things I could try?

Answer 1

Uniform convergence is a sufficient condition for sure. Sometimes Dini's theorem is helpful https://en.wikipedia.org/wiki/Dini%27s_theorem

Answer 2

Let $[a,b] = [0,1]$, with $f_n(x)$ given piecewise: for $x$ such that $0 \leq \frac{1}{2} - \frac{1}{2n}$, $f_n(x) = 0$. For $x$ such that $\frac{1}{2} - \frac{1}{2n} \leq x \leq \frac{1}{2}$, $f_n(x) = 2n^2x$, on $\frac{1}{2} \leq x \leq \frac{1}{2} + \frac{1}{2n}$, $f_n(x) = 1 - 2n^2x$, and on $\frac{1}{2} + \frac{1}{2n} \leq x \leq 1$, $f_n(x) = 0$.

The picture for this is a peak in the middle of the interval whose base gets narrower as it grows, and $\int_0^1 f_n(x)dx= 1$ for all $n$, but $f_n$ converges to the zero function away from $\frac{1}{2}$, although not uniformly, so in this case, the limit of the integrals is not the integral of the limit.