# Convergence of Integrals implies almost everywhere convergence of functions

I'm wondering whether the following statement holds:

Let $$f_n,f\colon \mathbb{R} \to \mathbb{R}^+_0$$ be functions with $$\int f_n(x) dx = \int f(x) dx=1$$ and for every bounded and countinuous function $$g\colon \mathbb{R} \to \mathbb{R}$$ the following integral-convergence $$\int g(x) \cdot f_n(x) dx \rightarrow_n \int g(x) \cdot f(x) dx$$ 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

A classical counterexample is $$f_n(x):=\bigl(1+\sin(2\pi nx)\bigr)\mathbf1_{\{x\in(0,1)\}},$$ and $$f(x):=\mathbf1_{\{x\in(0,1)\}}.$$ We have $$\int f_n=\int f=1$$ and, for every bounded continuous $$g\colon\mathbb R\to\mathbb R$$, $$\int f_ng\to\int fg$$ as $$n\to\infty$$ (e.g., by Riemann–Lebesgue lemma). However, $$(f_n)_{n\ge1}$$ does not converge at all.
• Could you clarify the reasoning meant by Riemann-Lebesuge lemma? We identify $\sin(2\pi nx) \cdot \mathbb{1}_{\lbrace x \in (0,1) \rbrace}$ as a Fourier-transform of a $L^1$ function, thus it converges to 0 for $n \to \infty$? Dec 6, 2021 at 22:21
• @fix foxi: Riemann-Lebesgue lemma gives you directly that $\int_0^1g(x)\sin(2\pi nx)\,\mathrm dx\to0$ as $n\to\infty$. Dec 6, 2021 at 22:26
• I may be tired, but $1+\sin(2\pi nx)$ is always $\ge0$, so $(1+\sin(2\pi nx))\mathrm e^{-x^2/2}/\sqrt{2\pi}\ge0$ for any $x$… Dec 6, 2021 at 23:04
• Hint. The integral of an odd, integrable function is $0$. ;) Dec 6, 2021 at 23:11