Riemann integral of product of sequence of function with a continuous function at a point. Here is the setup to a questions I've been looking at. I mainly just need a hint on how to proceed from where I'm at. Or suggest a theorem/corollaries that would help.
The problem:
Let $\{g_n\}$ be a sequence of nonnegative Riemann integrable function on $[-1,1]$ satisfying

*

*$\int_{-1}^{1} g_n(x) dx = 1$ for $n\in\mathbb{N}^+$

*For every $\delta>0$, $g_n \to 0$ uniformly on $[-1,-\delta] \cup [\delta, 1]$ as $n\to \infty$.

Prove that if $\psi:[-1,1] \to \mathbb{R}$ is Riemann integrable and continuous at $x=0$, then
$$
\lim_{n\to \infty} \int_{-1}^{1} \psi(x)g_n(x)dx = \psi(0).
$$
Attempt at solution
So by the properties of the integral and limit, and uniform convergence of $g_n$ we have that
$$
\lim_{n\to \infty} \int_{-1}^{1} \psi(x)g_n(x) dx = \lim_{n\to \infty} \int_{-1}^{-\delta} \psi(x)g_n(x) dx + \lim_{n\to \infty} \int_{-\delta}^{\delta} \psi(x)g_n(x) dx + \lim_{n\to \infty} \int_{\delta}^{1} \psi(x)g_n(x) dx \\
=\lim_{n\to \infty} \int_{-\delta}^{\delta} \psi(x)g_n(x) dx.
$$
Question
Am I correct in the above computations?
If so then it is clear here that I have to use condition "1" along with continuity of $\psi$ at $x=0$. However, this is where I'm entirely lost and not sure how to approach the integral
$$
\lim_{n\to \infty} \int_{-\delta}^{\delta} \psi(x)g_n(x) dx
$$
 A: Since $\psi$ is continuous at $x=0$, for any $\epsilon>0$, there is a $\delta>0$ that for any $|x|<\delta$, $\:|\psi(x)-\psi(0)|<\epsilon$. Also, since $\psi$ is Riemann integrable, on $[-1,-\delta] \cup [\delta, 1]$, $|\psi(x)-\psi(0)|<M$. Thus
\begin{align}
\left|\int_{-1}^{1} \psi(x)\,g_n(x)\,dx-\psi(0)\right|&=\left|\int_{-1}^{1} \psi(x)\,g_n(x)\,dx-\int_{-1}^{1} \psi(0)\,g_n(x)\,dx\right|
\\
&\leqslant\int_{-1}^{1} |\psi(x)-\psi(0)|\,g_n(x)\,dx
\\
&=\int_{-1}^{-\delta} |\psi(x)-\psi(0)|\,g_n(x)\,dx+\int_{-\delta}^{\delta} |\psi(x)-\psi(0)|\,g_n(x)\,dx+\int_{\delta}^{1} |\psi(x)-\psi(0)|\,g_n(x)\,dx
\\
&<M\int_{-1}^{-\delta} g_n(x)\,dx+ \epsilon \int_{-\delta}^{\delta} g_n(x)\,dx+M\int_{\delta}^{1} g_n(x)\,dx
\\
&<\epsilon + M\int_{-1}^{-\delta} g_n(x)\,dx+ M\int_{\delta}^{1} g_n(x)\,dx
\end{align}
Since $g_n \to 0$ uniformly on $[-1,-\delta] \cup [\delta, 1]$, there is a $N$ that for any $n>N,$
$$
\int_{-1}^{-\delta} g_n(x)\,dx<\epsilon/M\quad\text{and}\quad \int_{\delta}^{1} g_n(x)\,dx<\epsilon/M
$$
Thus
$$
\left|\int_{-1}^{1} \psi(x)\,g_n(x)\,dx-\psi(0)\right|<3\epsilon\quad\text{and}\quad\lim_{n\to \infty}\,\int_{-1}^{1} \psi(x)\,g_n(x)\,dx=\psi(0)
$$
