Prove that $\displaystyle \lim_{n\to\infty}\underline{\int_a^b}f_n(x)\ \text{d}x=0$ If $\{fn\}$ is a decreasing sequence of bounded functions in $[a, b]$ and if $$\lim_{n\to\infty}f
_n(x)=0$$
then $$ \lim_{n\to\infty}\underline{\int_a^b}f_n(x)\ \text{d}x=0$$
Any help or any suggestions to solve this exercise. Thanks a lot :).
Try: I have a little idea, which I don't know if it's correct. Notice that, $$\left |\underline{\int_a^b} f
_n(x)\, \text{d}x\right |\le \underline{\int_a^b}|f_n(x)|\, \text{d}x$$
If we take limits on both sides, we have that the right part goes to 0, and it remains that the absolute value is less than or equal to 0, the only possibility is that it is 0, and it would already have what we want.
 A: Here is a solution using either the monotone convergence theorem or the dominated convergence theorem. The trick is that the lower Darboux integral can be replaced by the Lebesgue integral of a certain function.
1. For each bounded function $ f : [a, b] \to \mathbb{R}$, define the function $\underline{f} : [a, b] \to \mathbb{R}$ as
$$ \underline{f}(x) = \sup\{ \varphi(x) : \varphi \in C[a,b] \text{ and } \varphi \leq f \}. $$
This is called the lower semicontinuous envelope of $f$, and it is measurable because, for each $y \in \mathbb{R}$
\begin{align*}
\underline{f}^{-1}((y, \infty))
&= \{x \in [a, b] : \underline{f}(x) > y \} \\
&= \{x \in [a, b] : \varphi(x) > y \text{ for some } \varphi \in C[a, b] \text{ s.t. } \varphi \leq f \} \\
&= \bigcup_{\substack{\varphi \in C[a, b] \\ \varphi \leq f}} \varphi^{-1}((y, \infty))
\end{align*}
is open in $[a, b]$.
2. Note that for each $\varepsilon > 0$, we can find $\varphi \in C[a, b]$ such that
$$ \varphi \leq f \qquad \text{and}\qquad \mathop{\underline{\int_{a}^{b}}} f(x) \, \mathrm{d}x - \int_{a}^{b} \varphi(x) \, \mathrm{d}x < \varepsilon. $$
So by choosing a sequence $(\varphi_n)_{n\in\mathbb{N}}$ in $C[a,b]$ such that $\varphi_1 \leq \varphi_2 \leq \cdots \leq f$ and $\int_{a}^{b} \varphi_n(x) \, \mathrm{d}x$ converges to $\mathop{\underline{\int_{a}^{b}}} f(x) \, \mathrm{d}x$, the squeezing lemma applied to $\varphi_n \leq \underline{f} \leq f$ shows that
$$ \mathop{\underline{\int_{a}^{b}}} f(x) \, \mathrm{d}x = \int_{a}^{b} \underline{f}(x) \, \mathrm{d}x. $$
3. Now we are ready. Let $(f_n)_{n\in\mathbb{N}}$ be as in the assumption. Then

*

*$(\underline{f}_n)_{n\in\mathbb{N}}$ is also a decreasing sequence of bounded, non-negative, measurable functions,


*$0 \leq \underline{f}_n \leq f_n$ shows that $\underline{f}_n(x) \downarrow 0$ as $n\to\infty$ for each $x \in [a, b]$.
So by either of the aforementioned convergence theorem, it follows that
\begin{align*}
\lim_{n\to\infty} \underline{\int_{a}^{b}} f_n (x) \, \mathrm{d}x
&= \lim_{n\to\infty} \int_{a}^{b} \underline{f}_n (x) \, \mathrm{d}x
= \int_{a}^{b} \lim_{n\to\infty} \underline{f}_n (x) \, \mathrm{d}x
= 0.
\end{align*}
A: https://sites.math.washington.edu/~morrow/335_17/dominated.pdf
A purely elementary proof can be found in Lemma 2.2 in the above.
The key ideas are to get continuous functions which are less than $f_i(x)$ but which have similar integral using the definition of the lower integral and smoothing the indicator functions, and taking appropriate minima. Then, these continuous functions are uniformly continuous, so in the limit, their integral and hence $f$’s lower integral goes to 0.
