Convergence locally uniformly VS $L^1$ convegence for probability density functions 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.
 A: 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 $L^1(0,\infty)$.
