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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

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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.

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  • $\begingroup$ 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 $n \to \infty$? $\endgroup$
    – barri
    Dec 6, 2021 at 22:21
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    $\begingroup$ @fix foxi: Riemann-Lebesgue lemma gives you directly that $\int_0^1g(x)\sin(2\pi nx)\,\mathrm dx\to0$ as $n\to\infty$. $\endgroup$
    – nejimban
    Dec 6, 2021 at 22:26
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    $\begingroup$ 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 $x$… $\endgroup$
    – nejimban
    Dec 6, 2021 at 23:04
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    $\begingroup$ Hint. The integral of an odd, integrable function is $0$. ;) $\endgroup$
    – nejimban
    Dec 6, 2021 at 23:11
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    $\begingroup$ Just realized that. Thank you very much for the help and solution of the question! $\endgroup$
    – barri
    Dec 6, 2021 at 23:12

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