Let $f_1, f_2, \ldots$ and $f$ be probability density functions on $(0, \infty)$ - so $\int_{(0,\infty)}f_n(x)dx=1$, $\int_{(0,\infty)}f(x)dx=1$. Assume that for every $x \in (0, \infty)$ there exists a neighborhood $N_x$ of $x$ such that
$$ \sup_{y \in N_x}|f_n(y)-f(y)|\to 0, $$
i.e. $f_n$ converges to $f$ locally uniformly. This can also be thought in terms of uniform convergence on compact sets. Now, is it possible to also conclude that $\Vert f_n - f \Vert_1 \to 0$? My feeling is that the answer is no, unless additional assumptions are included, but I'm not managing to construct a counterexample.
Yes, it is possible to conclude that the convergence is in $L^1$ norm.
Given any $\varepsilon>0$, there exists a compact set $K$ such that $$ \int_{(0,\infty)\backslash K} f(x)\, dx<\varepsilon/4 $$ since $f\in L^1(0,\infty)$. On the other hand, uniform convergence on $K$, a bounded set, implies $L^1(K)$ convergence, hence $\exists N\in\Bbb N$ such that $$ \int_{ K} |f(x)-f_n(x)| \, dx <\varepsilon/4 $$ for all $n\ge N$. This also implies that for each $n\ge N$, we have $$\begin{align} \int_{ K} |f_n(x) |\, dx &\ge \int_{ K} |f(x)|\, dx -\int_{ K} |f(x)-f_n(x)| \, dx \\ &\ge (1-\varepsilon/4) - \varepsilon/4 \\ &= 1 - \varepsilon/2, \end{align}$$ hence $$ \int_{(0,\infty)\backslash K} |f_n(x) |\, dx \le \varepsilon/2. $$ This means that for all $n\ge N$, $$\begin{align} \int_{(0,\infty)} |f(x)-f_n(x)| \, dx &\le \int_{K} |f(x)-f_n(x)| \, dx + \int_{(0,\infty)\backslash K} |f(x)-f_n(x)| \, dx \\ &\le \varepsilon/4 + \int_{(0,\infty)\backslash K} |f(x)| \, dx +\int_{(0,\infty)\backslash K} |f_n(x)| \, dx \\ &\le \varepsilon/4 + \varepsilon/4 + \varepsilon/2 \\ &\le \varepsilon, \end{align}$$ which proves convergence in $L^1(0,\infty)$.