Let $f$ be (Riemann) integrable over $[0,1]$. Show $\sum_{n=0}^{\infty}\int_{0}^{x^n}f(t)dt$ is continuous on $(0,1)$

Question

Problem: Let $f$ be (Riemann) integrable over $[0,1]$. Show $\sum_{n=0}^{\infty}\int_{0}^{x^n}f(t)dt$ is continuous on $(0,1)$.

I know that if $f$ is (Riemann) integrable over $[0,1]$, since $x \in (0,1)$ then $x^n \in (0,1)$ also and every integral exists. Clearly the integral eventually goes to $0$ but I'm not sure how to use that. Also, I know that given an $f(t)$, I can use the epsilon-delta definition of continuity to show it's continuous, but here I'm only given that it's Riemann integrable. Would I do something like show that $\int_{0}^{x^n}f(t)dt$ is bounded, and use the epsilon-delta to show the whole thing must be continuous (as in Continuity of function consisting of an infinite series. for example)? I'm not sure how to do this rigorously though. Any help would be much appreciated!

Answer 1

The function $$F(x):=\int_0^x f(t)dt$$ is Lipschitz, as one easily sees, since $|F(x)|\le \int_0^x|f|\le x\underset{[0,1]}{\text{sup}}(|f|)$. Let $K$ be its Lipschitz constant. Then

$$\sum_{n=0}^\infty\left|\int_0^{x^n}f(t)dt\right|=\sum_{n=0}^{\infty}|F(x^n)|\le K\sum_{n=0}^\infty x^n=K\frac{1}{1-x}$$

Since the series is locally normally convergent and every term of the series is continuous, it is continuous.

One may ask if the method can be pushed further, i.e. if we can try to prove that the series is continuous also in $1$: this is not the case. Take $f(x)=x$ to find a counterexample.

Answer 2

Let $t \in (0, 1)$. We first show that the series is continuous on $[0, t]$. First, note that $f$ must be bounded on $[0, 1]$ since it is Riemann integrable. Let $S = \sup |f|$.

We show that the sum converges uniformly on $[0, t]$. For this, we use Cauchy's criteria. That is, given any $\epsilon > 0$, there exists $n \in \mathbb{N}$ such that for all $M \ge N \ge n$, we have

$$\sup_{x \in [0, t]}\left|\sum_{k=N}^M\int_0^{x^k}f(t)dt\right| \le \epsilon.$$

Of course, it suffices to show that the LHS of the above inequality can be made arbitrarily small.

For $x \in [0, t]$, we have

$$\left|\sum_{k=N}^M\int_0^{x^k}f(t)dt\right| \le \sum_{k=N}^M\int_0^{x^k}Sdt = S\sum_{k=N}^Mx^k \le S\sum_{k=N}^Mt^k.$$

The last summation can be made arbitrarily small since $\sum t^n$ converges. ($\because 0 < t < 1.$)

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Thus, we have that $F(x) = \displaystyle\sum_{n = 0}^\infty\int_0^{x^n}f(t)dt$ is well-defined on $[0, 1)$ and is continuous on $[0, t]$ for every $t < 1$.

Now, we show that $f$ is continuous on $[0, 1)$. Let $x \in [0, 1)$. We show that $f$ is continous at $x$.
Choose $t$ such that $x < t < 1$. Then, $f$ is continuous on $[0, t]$ hence, continuous at $x$ and thus, we are done.