Does convergence on compacta imply $L_{1}$-convergence?

Question

Let $G$ be a locally compact Hausdorff group with a Haar measure $\mu$. and let $(f_{n})$ be a sequence in $C_{c}(G)$ that converges uniformly to a continuous function $f\in L^{1}(G)\cap C(G)$ on every compact subset, that is, $$\|f|_{K}-f_{n}|_{K}\|_{\infty}\to0$$ for every compact subset $K\subset G$. Does this imply that $(f_{n})$ converges to $f$ with respect to the $L_{1}$-norm to $f$, that is, $$\|f-f_{n}\|_{1}=\int_{G}|f-f_{n}| \ \text{d}\mu\to0?$$ This is clearly true if $G$ is compact, since $$\|f-f_{n}\|_{1}\leq\mu(G)\cdot\|f-f_{n}\|_{\infty}.$$

Answer 1

No, take $G=(\mathbb R,+)$. Then the Haar measure is a positive multiple of the Lebesgue measure. Let $f_n$ be continuous and such that $f_n = 1/n$ on $[-n,+n]$ and $f_n(x)=0$ for all $x$ with $|x|\ge n+1$. Then $f_n \to0$ uniformly, but $\|f_n\|_{L^1} \ge 2$ for all $n$.