The sequence of the continuous functions $f_n$ on the $I= [-1,1]$ satisfies $(a)$~$(c)$ [Here $\mathbb{N}$ is the natural number set]
$(a)$ $f_n(x) \geq 0 $ for $\forall x \in I$, $\forall n \in \mathbb{N}$
$(b)$ $\forall n \in \mathbb{N}$ , $\int_{-1}^{1}f_n(x) dx =1$
$(c)$ For all $a$ which is $0 < a <1$, $\lim\limits_{n \to \infty} \int_{a \leq \vert x \vert \leq 1}f_n(x)dx=0 $
Put the $g_n(x) = \int_{-1}^{1}f_n(t)e^{x-t}dt $ which is $g_n : [0,1]\to \mathbb{R}$. Find the value of $\lim\limits_{n \to \infty} \int_{0}^{1}g_n(x)dx$.
In solution sheet, It found the value by using epsilon-delta technique. It used the definition of the uniformly convergence. But I tried different ways like the below.
From the $(b)$, $1= \lim\limits_{n \to \infty}\int_{-1}^{1}f_n(x) dx = \lim\limits_{n \to \infty}\int_{-\frac{1}{n}}^{\frac{1}{n}}f_n(x) dx + \lim\limits_{n \to \infty}\int_{\frac{1}{n}\leq\vert x \vert \leq 1}f_n(x) dx$.
Then By $(c)$, $1= \lim\limits_{n \to \infty}\int_{-\frac{1}{n}}^{\frac{1}{n}}f_n(x) dx $
The question requires $\lim\limits_{n \to \infty} \int_{0}^{1}g_n(x)dx = (e-1) \{\int_{-\frac{1}{n}}^{\frac{1}{n}}f_n(x)e^{-x} dx + \int_{\frac{1}{n}\leq\vert x \vert \leq 1}f_n(x)e^{-x} dx \}$.
First, By the mean value thm of the integral, I got $ \int_{-\frac{1}{n}}^{\frac{1}{n}}f_n(x)e^{-x} dx = e^{-{c_n}}\int_{-\frac{1}{n}}^{\frac{1}{n}}f_n(x) dx$ for the $c_n\in (- \frac{1}{n}, \frac{1}{n})$
So I deduced $1= \lim\limits_{n \to \infty}\int_{-\frac{1}{n}}^{\frac{1}{n}}f_n(x)e^{-x} dx $.
Second from the $(a)$, $e^{-1} \int_{\frac{1}{n}\leq\vert x \vert \leq 1}f_n(x) dx \leq \int_{\frac{1}{n}\leq\vert x \vert \leq 1}e^{-x}f_n(x) dx \leq e\int_{\frac{1}{n}\leq\vert x \vert \leq 1}f_n(x) dx $ for the $x \in [-1,1]$
By squeeze thm and $(c)$, $\lim\limits_{n \to \infty}\int_{\frac{1}{n} \leq \vert x \vert \leq 1}f_n(x)e^{-x} dx = 0$ .
When we back to the point of the $g_n$, the value would be $(e-1)$ considering the first and second arguments.
Is my solution is right? If not please let me know which point I've missed. Thank you.