Let the sequence of the continuous functions, $f_n : [0,1] \to \mathbb{R}$ with $f_n(x) \geq 0$ on $[0,1]$
The $f_n$ has the below properties.
(1) $\int_0^1 f_n(x) dx = 1$ for all $n \in \mathbb{N}$
(2) $f_n(x)$ is pointwise convergent to $0$ for $\forall x \in [0,1]$
(3) $\forall r \in \mathbb{R} s.t. 0 \lt r \lt 1, f_n(x)$ is uniformly convergent on $[r,1]$
Say the $g(x) = f_n(x)e^{-x}$, What is $lim_{n \to \infty} $$\int_{0}^{1}g(x)dx$ ?
My attempt) By property (3)
$lim_{n \to \infty}$ $\int_0^1g(x)dx$ = $lim_{n \to \infty}$ ( $\int_0^rg(x)dx$ + $\int_r^1g(x)dx$ ) = $lim_{n \to \infty}$ $\int_0^rg(x)dx$
Then by M.V.T. for the integral
$\exists c s.t. 0 \lt c \lt r \lt 1$, $lim_{n \to \infty}$ $e^{-c} \int_0^r f_n(x)dx$
As $r \to 0$, clearly $c \to 0$
Hence the limit is $0$
But the answer was $1$.
I can't figure out which point do I have a mistake. Plus in my answer sheet, There are no method finding the value of the limit is $1$. Still I don't know why the value is $1$. Any help would be appreciated. Thanks.
You did not prove that the limit is $0$. Here is a valid argument:
Consider $\int_0^{1} [1-e^{-x}] f_n(x)dx$. Using the inequality $1-e^{-x} \leq x$ we see that $$\int_0^{r} [1-e^{-x}] f_n(x)dx \leq \int_0^{r} xf_n(x)dx \leq r \int_0^{r} f_n(x)dx \leq r.$$ Also $\int_r^{1} [1-e^{-x}] f_n(x)dx \to 0$ by uniform convergence. Conclude that $\int_0^{1} [1-e^{-x}] f_n(x)dx \to 0$ which is equivalent to the statement $\int_0^{1} g_n(x)dx \to 1$.