Uniform convergence of continuous functions $f_n$ to $f >0$ a.e., implies convergence of Riemann sum involving $1_{(-\infty , 0)}(f_n(s))$ Let $f :[0,T] \to \Bbb R$ be a continuous function with $\int_0^T 1_{f^{-1}(0)} (s) ds = 0$.
Let $f_n \to f$ uniformly on $[0,T]$.
Then $$\sum_{i=1}^n \frac T n 1_{(-\infty , 0)}\left(f_n\left(T \cdot \frac i n\right)\right) \to \int_0^T 1_{(-\infty , 0)} (f(s))ds .$$
Question: Is there a shorter proof than the one below or even a reference I could directly employ to this problem?
proof: For $\varepsilon > 0$ let $\phi_\varepsilon^1$ be a Lipschitz-continuous function with $\phi_\varepsilon^1 (x) = 1$ for $x\leq -\varepsilon$, $\phi_\varepsilon^1 (x) = 0$ for $x\geq \varepsilon$ , $0\leq \phi_\varepsilon^1 \leq 1$ and $\phi_\varepsilon^2 $ a Lipschitz-continuous function with $\phi_\varepsilon^2 (x) = 0$ for $\vert x \vert > 2\varepsilon$, $\phi_\varepsilon^2 (x) = 1$ for $\vert x \vert \leq \varepsilon$, $0\leq \phi_\varepsilon^2\leq 1$. Then
$$\vert \sum_{i=1}^n \frac T n \phi_\varepsilon^k(f_n( iT/n)) - \sum_{i=1}^n \frac T n \phi_\varepsilon^k(f( iT/n)) \vert \to 0$$
by the Lipschitz condition and the uniform convergence and
$$\vert \sum_{i=1}^n \frac T n \phi_\varepsilon^1(f( iT/n)) - \sum_{i=1}^n \frac T n 1_{(-\infty , 0)}(f( iT/n)) \vert\\
\leq \sum_{i=1}^n \frac T n \phi_\varepsilon^2(f( iT/n)) \to \int_0^T \phi_\varepsilon ^2(f(s) )ds \to 0$$
as $\varepsilon \to 0$ by dominated convergence and $\int_0^T 1_{f^{-1}(0)} (s) ds=0$. Furthermore
$$\vert \sum_{i=1}^n \frac T n \phi_\varepsilon^1(f_n( iT/n)) - \sum_{i=1}^n \frac T n 1_{(-\infty , 0)}(f_n( iT/n)) \vert \\
\leq \sum_{i=1}^n \frac T n \phi_\varepsilon^2(f_n( iT/n)) \to \int_0^T \phi_\varepsilon ^2(f(s) )ds \to 0$$
as $\varepsilon \to 0$ by dominated convergence and $\int_0^T 1_{f^{-1}(0)} (s) ds=0$. Hence
$$\lim_{n\to\infty}\sum_{i=1}^n \frac T n 1_{(-\infty , 0)}(f_n( iT/n)) = \lim_{\varepsilon \to 0}\lim_{n\to\infty} \sum_{i=1}^n \frac T n \phi_\varepsilon^1(f( iT/n)) \\= \lim_{\varepsilon \to 0}\int_0^T \phi_\varepsilon ^1(f(s) )ds = \int_0^T 1_{(-\infty , 0)} (f(s))ds$$
again by dominated convergence and $\int_0^T 1_{f^{-1}(0)} (s) ds =0$.
 A: By the following one can prove the statement a bit shorter:
By the fact that $\{t : s\mapsto 1_{(-\infty , 0)}(f(s))\} \subseteq f^{-1} (0)$ and $\int_0^T 1_{f^{-1} (\{0\})} (s) ds = 0$ we have by Lebesgue's criterion for Riemann integrability that
$$\sum_{i=1}^n \frac T n 1_{(-\infty , 0)}(f( iT/n)) \to \int_0^T 1_{(-\infty , 0)} (f(s))ds$$
Further we have for any $\varepsilon >0$ and $n$ large enough such that $\sup_{t\in[0,T]}\vert f_n(t)-f(t)\vert < \varepsilon$ that
$$\sum_{i=1}^n \frac T n \Big\vert 1_{(-\infty , 0)}(f_n( iT/n)) - 1_{(-\infty , 0)}(f( iT/n)) \Big \vert \\
=\sum_{i=1}^n \frac T n 1_{(-\infty , 0)}(f_n( iT/n))1_{(0,\infty)}(f( iT/n)) + 1_{(0,\infty)}(f_n( iT/n)) 1_{(-\infty , 0)}(f( iT/n)) \\
=\sum_{i=1}^n \frac T n 1_{(-\varepsilon, 0)}(f_n( iT/n))1_{(0,\varepsilon)}(f( iT/n)) + 1_{(0,\varepsilon)}(f_n( iT/n)) 1_{(-\varepsilon , 0)}(f( iT/n)) \\
\leq \sum_{i=1}^n \frac T n 1_{(-\varepsilon, \varepsilon)}(f( iT/n)) \leq \sum_{i=1}^n \frac T n \phi_\varepsilon^2(f( iT/n)) \\
\overset{n\to\infty}{\to} \int_0^T \phi_\varepsilon^2 (f(s))ds \overset{\varepsilon \to 0}{\to} \int_0^T 1_{f^{-1}(\{0\})} (s) ds = 0,
$$
which proves the statement.
