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?

  • $\begingroup$ Why not using DNT? $\endgroup$
    – vitamin d
    Mar 23, 2021 at 22:27
  • $\begingroup$ @vitamind we have not covered dominated convergence theorem. We cannot use it. Sorry about that. $\endgroup$
    – Nolan P
    Mar 23, 2021 at 22:28

2 Answers 2


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.

  • $\begingroup$ I can't really write as a proof "look at the graph and you'll see it." My professor would hate that. I need to actually prove it. How can I do so? $\endgroup$
    – Nolan P
    Mar 23, 2021 at 22:29
  • $\begingroup$ Consider $y = xe^{-x}$ for $x \geq 0$, and maximise it. How would you normally maximise a function of a single variable like this? $\endgroup$
    – JKL
    Mar 23, 2021 at 22:30
  • $\begingroup$ I would use the first derivative test as we do in Calc I. Take the derivative, set it equal $0$, and proceed. $\endgroup$
    – Nolan P
    Mar 23, 2021 at 22:32
  • $\begingroup$ That seems like a sound strategy - then that would show that $x e^{-x} \leq M$ for some $M > 0$, and this holds for any $x > 0$. $\endgroup$
    – JKL
    Mar 23, 2021 at 22:34
  • $\begingroup$ Ah because I can find a maximum and then of course I have a bound. Now, can I do this in general for the function $f_n(x)$? It would be $0 = f'_n(x)e^{-f_n(x)} - f'_n(x)f_n(x)e^{-f_n(x)}$ I can factor and make this $0 = f'_n(x)e^{-f_n(x)}(1 - f_n(x))$. If the term I factored out was $0$, we would be done trivally. Considering the second term, I'd have $f_n(x) = 1$. Is that my bound? $\endgroup$
    – Nolan P
    Mar 23, 2021 at 22:42

$e^{t} \geq t$ 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 $t \geq 1$.


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