So I have a sequence of riemann (darboux specifically) integrable real valued functions $(f_n)$ defined on $[a,b] \subset \mathbb{R}$ that converges uniformly to some $f$. I would like a hint (or two even) on how to show this $f$ is integrable. I split this problem up into two parts, namely first proving that $f$ is bounded, and then later proving that $f$ is integrable. So for the first part I have that:

Recall that each $f_n$ is bounded on $[a,b]$, so we have that for each $n$, $f_n([a,b])$ has some supremum in $\mathbb{R}$ (call it $M$) and that: $$ f_n(x) \leq M \text{ for any $x\in [a,b]$} $$ A similar statement can be made about the lower bound. Fix some $\epsilon>0$. We know by the uniform convergence of $f_n$ to $f(x)$ that, there exists $N \in \mathbb{N}$ with the property that for any $n \geq N$ we have: $$ |f_n(x) - f(x)|<\epsilon \text{ for all $x\in [a,b]$} $$ So we see that: $$ f(x)< f_n(x) + \epsilon \leq M +\epsilon \text{ for all $x\in [a,b]$} $$

So $f$ is bounded.

I am not entirely sure how to proceed from here. I am fairly certain I need to satisfy the Riemann integrability condition, i.e show that for any $\epsilon$ there exists a partition $P$ of $[a,b]$ such that: $$ U(f,P) - L(f,P) < \epsilon $$ I have a feeling I need to pick some special epsilons and do some tricky manipulations, but I was hoping for some guidance in the right direction.

  • $\begingroup$ You are correct that the way to go here is to just do the manipulations. Thankfully, it's not too bad, since the uniform convergence will let you estimate what happens with the partition in a reasonably nice way. $\endgroup$ – Alfred Yerger Apr 27 '17 at 5:23

Some care is required in estimating $U(f,P) - L(f,P)$.

By uniform convergence, if $n$ is sufficiently large, then

$$-\frac{\epsilon}{3(b-a)} < f(x) - f_n(x) < \frac{\epsilon}{3(b-a)}.$$

We have

$$f(x) = f(x) - f_n(x) + f_n(x),$$

implying, on any partition subinterval $I$,

$$\sup_I f(x) \leqslant \sup_I(f(x) - f_n(x)) + \sup_I f_n(x) < \frac{\epsilon}{3(b-a)}+ \sup_I f_n(x), \\ \inf_I f(x) \geqslant \inf_I(f(x) - f_n(x)) + \inf_I f_n(x) > -\frac{\epsilon}{3(b-a)}+ \inf_I f_n(x).$$

The second chain of inequalities implies

$$-\inf_I f(x) < \frac{\epsilon}{3(b-a)} - \inf_I f_n(x).$$

Summing over all partition subintervals we get

$$U(f,P) < \frac{\epsilon}{3} + U(f_n,P),\\ -L(f,P) < \frac{\epsilon}{3} - L(f_n,P).$$


$$U(f,P) - L(f,P) < \frac{2\epsilon}{3} + U(f_n,P) - L(f_n,P).$$

Since $f_n$ is Riemann integrable, there is a partition $P$ such that $U(f_n,P) - L(f_n,P) < \epsilon/3$ and the desired result $U(f,P) - L(f,P) < \epsilon$ follows.


HINT: Write \begin{equation*} \begin{aligned} U(f,P) - L(f,P) &= U(f,P) - U(f_n,P) + U(f_n,P) - L(f_n,P) + L(f_n,P) - L(f,P) \\ &\leq \vert U(f,P) - U(f_n,P) \vert + U(f_n,P) - L(f_n,P) + \vert L(f_n,P) - L(f,P) \vert. \end{aligned} \end{equation*}

As an aside, it's worth noting that once you prove this, you can also use uniform continuity to prove that $$\lim_{n\to\infty}\int_a^b f_n(x)dx = \int_a^b f(x)dx,$$ i.e. you can interchange the order of the limit and the integral.


You may not like this method, but here goes!

A function is Riemann integrable on $[a,b]$ iff it's bounded, Lebesgue measurable and continuous almost everywhere in the Lebesgue sense. Boundedness and Lebesgue measurability are preserved under uniform limits. As a countable union of measure zero sets has measure zero, then it suffices to prove that $f$ is continuous at each $x\in[a,b]$ at which all the $f_n$ are continuous. The usual "$3\varepsilon$ argument" for continuity of uniform limits works for this.

  • $\begingroup$ I am sure this answer is correct but I was told the proof could be done without any measure theoretic concepts. $\endgroup$ – rubikscube09 Apr 27 '17 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.