 Riya Hansen

2022-07-03

I'm wondering whether the following statement holds:
Let ${f}_{n},f:\mathbb{R}\to {\mathbb{R}}_{0}^{+}$ be functions with $\int {f}_{n}\left(x\right)dx=\int f\left(x\right)dx=1$ and for every bounded and countinuous function $g:\mathbb{R}\to \mathbb{R}$ the following integral-convergence
$\int g\left(x\right)\cdot {f}_{n}\left(x\right)dx{\to }_{n}\int g\left(x\right)\cdot f\left(x\right)dx$
holds. Then it follows that ${f}_{n}\to f$ almost everywhere.
Intuitively the statement looks false, but I can't find a counterexample. If it doesn't hold: changes the further assumption that the ${f}_{n},f$ have to be continuous anything?
Kind regards Franco Cohen

Expert

A classical counterexample is
${f}_{n}\left(x\right):=\left(1+\mathrm{sin}\left(2\pi nx\right)\right){\mathbf{1}}_{\left\{x\in \left(0,1\right)\right\}},$
and
$f\left(x\right):={\mathbf{1}}_{\left\{x\in \left(0,1\right)\right\}}.$
We have $\int {f}_{n}=\int f=1$ and, for every bounded continuous $g:\mathbb{R}\to \mathbb{R}$, $\int {f}_{n}g\to \int fg$ as $n\to \mathrm{\infty }$ (e.g., by Riemann–Lebesgue lemma). However, $\left({f}_{n}{\right)}_{n\ge 1}$ does not converge at all. Bruno Pittman

Expert

Riemann-Lebesgue lemma gives you directly that ${\int }_{0}^{1}g\left(x\right)\mathrm{sin}\left(2\pi nx\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\to 0$ as $n\to \mathrm{\infty }$