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Let $f_n$ be a sequence of Riemann integrable functions on $[a,b]$ that converge uniformly to a function f. Show that f is also Riemann integrable. What happens if $f_n$ only converges pointwise?

Considering this scenerio, show that

$$\lim_{n \to \infty} \int_{a}^{b} f_n(x) dx = \int_{a}^{b} f(x) dx$$

Not really sure how to start this problem. If they already converge to a function, uniformly, that function has to be continuous right? So then it is trivially Riemann integrable. The pointwise case I'm not sure about. And then the second part with the limits I'm not sure how to go about. Any help is appreciated!

Nolan P
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2 Answers2

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Let $\{q_n\}_{n \in \mathbb{N}}$ be the rational numbers in the interval $[0,1]$, and let us consider the functions $$f_n(x)= \begin{cases} 1 &\text{if}\quad x \in \{q_1, \ldots, q_n\},\\ 0 &\text{otherwise.} \end{cases} $$

The $f_n(x)$ are Riemann-integrable but they converge to the Dirichlet function, which is not Riemann-integrable.

Bargabbiati
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We can use the Riemann criterion to prove that the uniform limit $f$ of a sequence of Riemann integrable functions $(f_n)_n$ is also Riemann integrable.

By uniform convergence, for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ we have

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

Let $P: a = x_0 < x_1 < \ldots < x_n = b$ be a partition. Since $f(x) = f(x) - f_n(x) + f_n(x),$ it follows that on any partition subinterval $I$,

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

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

Summing over all partition subintervals we get for upper and lower Darboux sums,

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

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

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

Now you should be able to prove on your own that the limit of the sequence of integrals is the integral of the limit function by considering that $|f_n(x) - f(x)| \to 0$ uniformly for all $x \in [a,b]$.

RRL
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  • This makes perfect sense, great explanation. For the second part, I'm thinking since you say $|f_n(x) - f(x)| \to 0$ as $n \to \infty$ would mean that the difference between these functions gets smaller and smaller showing they are relatively the same? Is that the arguement? I'm not so sure how I would show that second part yet. – Nolan P Dec 11 '20 at 17:23
  • Having proved first that $f$ is integrable, we have $\left|\int_a^b f_n(x) , dx - \int_a^bf(x) ,dx \right| \leqslant \int_a^b|f_n(x) - f(x)| ,dx$. Also by uniform convergence, there exists $N$ such that $n > N$ implies that $|f_n(x) - f(x)| < \frac{\epsilon}{b-a}$ for all $x \in [a,b]$. Do you see where this is going? – RRL Dec 11 '20 at 17:54
  • Ahhh yes I see it. That makes perfect sense. I will say, you do a fantastic job explaining things. Of all the questions I've asked on here your explanation of this one is probably the clearest I have gotten. I really appreciated it and keep doing what you're doing! – Nolan P Dec 11 '20 at 17:57
  • @DominicBlanco: You're welcome. – RRL Dec 11 '20 at 17:59