Pointwise convergence doesn't imply $L^p$ convergence if $p=\infty$ under some hypothesis

Question

I just proved that if $1\leq p<\infty$ and $f,f_n$ is a sequence of measurable functions such that $f_n(x)\rightarrow f(x)$ a.e $x\in X$ and $\exists g\in L^p(\mu)$ such that $|f_n(x)|\leq g(x)$ a.e $x\in X$, then $f_n\rightarrow f$ in $L^p.$

Why is it that this is not true if $p=\infty$?

Answer 1

When $p=+\infty$, the statement reads as follows:

If $\left(f_n\right)_{n\geqslant 1}$ is a sequence of measurable functions such that $f_n(x) \to f(x)$ almost everywhere and there exists a constant $M$ such that $\left|f_n(x)\right|\leqslant M$ for every $n$ and almost every $x$, then $\lVert f_n-f\rVert_{\mathbb L^{\infty}}\to 0$.

But we can imagine a sequence of functions taking the values $0$ and $1$, with pointwise but not uniform convergence. We choose for example $f_n(x):=1$ if $0\lt x\lt 1/n$ and $0$ otherwise (on the unit interval, with Borel $\sigma$-algebra and Lebesgue measure).