Show the following satisfies bounded convergence theorem

Question

Assume $f_n: [0,1] \to [0,\infty)$ is integrable for each $n$ and $(f_n)_{n = 1}^{\infty}$ converges pointwise a.e. to $f$. Prove that

$$\lim_{n \to \infty} \int_{[0,1]} f_n(x) e^{-f_n(x)}dx = \int_{[0,1]} f(x)e^{-f(x)}dx$$

I think I am very close. To satisfy the BND convergence theorem, we need the input to be of finite measure, which we have: $m([0,1]) = 1 - 0 = 1 < \infty$. We also need the sequence of functions to converge pointwise, which we have by assumption.

The only thing I believe I am missing is that $\exists c \in (0,\infty)$ s.t. $|f_n(x)| \leq c \ \forall n \in \mathbb{N}$. If I had this, then BND convergence theorem would apply.

Maybe I'm not seeing it, but what in this problem implies we have this bound? Can anyone point me in the right direction?

Answer 1

Note that $f_n(x) e^{-f_n(x)}$ is uniformly bounded (EDIT: Since the functions $f_n$ are non-negative).

(For example, a plot of $x e^{-x}$ will show it immediately.) So the bounded convergence theorem applies as you are on a space of finite measure, and the sequence of functions you are integrating are uniformly bounded.

Note that the bounded convergence theorem is a special case of the Dominated Convergence Theorem.

Answer 2

$e^{t} \geq t$ for all $t \geq 0$. Hence, $te^{-t} \leq 1$. Taking $t=f_n(x)$ it follows that $f_n(x)e^{-f_n(x)} \leq 1$ for all $n$ and all $x$.

Two proofs of the inequality $e^{t}\geq t$ for $t \geq 0$:

1. Use series expansion of $e^{t}$.

2. Let $f(t)=e^{t}-t$. Then $f'(t)=e^{t}-1 \geq 0$ so $f$ is non-decreasing. But $f(0)=1$ so $f(t) \geq 1 >0$ for all $t \geq 1$.