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.