Let $f: [0,1] \to \mathbb{R}$ be continuous, prove $\lim_{n\to \infty} \int_0^1 f(x^n)dx = f(0)$

This makes some sense looking at it. I have only the Regulated Integral definition of integration to work with :https://en.wikipedia.org/wiki/Regulated_integral

It uses sequences of step functions with a partition over a closed interval.

So I need to solve the integral prior to taking the limit but not sure how to really get what I need. Since $f $ is continuous function there exists a sequence $(\varphi_n)_{n \in \mathbb{N}}$ of step functions such that $\lim_{n \to \infty} \sup_{x \in [0,1]} \mid f(x) - \varphi_n(x)\mid \,= 0$. Moreover, $\int_0^1 f(x)dx := \lim_{m \to \infty} \int_0^1 \varphi_m(x)dx $ so it should be the case that $\int_0^1 f(x^n)dx := \lim_{m \to \infty} \int_0^1 \varphi_m(x^n)dx $ so I assume that it would be that $\lim_{n \to \infty } \int_0^1 f(x^n)dx := \lim_{n \to\infty} (\lim_{n \to \infty} \int_0^1 \varphi_m(x^n)dx )$

Now, $ \int_a^b \varphi(\eta)d\eta := \sum_{j=0}^{N}\varphi(\eta_j)(\sigma_{j+1}-\sigma_j)$ where $(\sigma_j)_{j=0}^{N+1})$ is a partition of $[a,b]$ and $\eta_{j} \in (\sigma_j,\sigma_{j+1})$ such that each block of the partition is constant so choice of $\eta_j$ is not particularly important.

So should have $ lim_{n \to \infty}(lim_{m \to \infty}\int_a^b \varphi_m(\eta^n)d\eta := lim_{n \to \infty}(\lim_{m \to \infty}\sum_{j=0}^{N}\varphi_m(\eta_j^n)(\sigma_{j+1}-\sigma_j))$.

This is where I get stuck.

  • 4
    $\begingroup$ Just to check: you are not familiar with the Dominated Convergence Theorem, I assume? (This would give a rather concise proof, even if it is overkill). $\endgroup$
    – Clement C.
    Jan 17, 2017 at 0:30
  • $\begingroup$ @ClementC. Man, I just love overkill! Kill the cockroaches with nukes by all means!!! $\endgroup$ Jan 17, 2017 at 0:38
  • $\begingroup$ @ClementC. No I am not, I will look it up but I can't use it if we haven't covered it yet. $\endgroup$ Jan 17, 2017 at 0:39
  • 1
    $\begingroup$ @oliverjones Without summoning the full artillery, you can also give a more basic $\varepsilon$-$\delta$ based argument -- see my answer. $\endgroup$
    – Clement C.
    Jan 17, 2017 at 0:52
  • 2
    $\begingroup$ Clement provided an excellent answer, but here's perhaps a less confusing argument: suppose WLOG that $f(0) = 0$. Using the change of variables $u=x^n$, we get: $$\frac1n \int_0^1 \frac{f(x)}{x^{1-\frac1n}} dx:=I(n)$$ Let $\epsilon >0$ and choose $\delta < 1$ such that $|f(x)|<\epsilon/2$ for $0\le x < \delta$. The quantity $\int_{\delta}^1 \frac{f(x)}{x^{1-\frac1n}} dx$ is absolutely bounded by a constant, so choose $N$ such that for $n\ge N$, $\frac1n \int_{\delta}^1 \frac{f(x)}{x^{1-\frac1n}} dx < \epsilon/2$. It follows that for $n\ge N$, $|I(n)| <\epsilon$. $\endgroup$
    – user384138
    Jan 17, 2017 at 0:54

3 Answers 3


As I mentioned in a comment, if you allow the Dominated Convergence Theorem then this is a matter of a few lines. The sequence of functions $(f_n)_n$ defined by $f_n(x) \stackrel{\rm def}{=} f(x^n)$ ($x\in[0,1])$ converges pointwise (by continuity of $f$) to the constant function $f(0)$ on $[0,1)$, and one can uniformly dominate all $f_n$'s by the integrable (because constant) function $g\stackrel{\rm def}{=} \lVert f\rVert_\infty$ ($f$ being bounded on $[0,1]$, since it is continuous). Now, by the DCT we get $$ \int_0^1 f_n \xrightarrow[n\to\infty]{} \int_0^1 \lim_{n\to\infty}f_n = \int_0^1 f(0)dx = f(0). $$

But this is overkill. Another option, explored below, would be to use continuity of $f$ at $0$: recall that $f$ being continuous on a closed interval, it is bounded.

Without loss of generality, we can assume $f(0) = 0$ (indeed, just apply the same exercise to $g\stackrel{\rm def}{=} f-f(0)$. It is easy to check that this will be sufficient to prove the result).

Fix $\varepsilon>0$, break the interval in 2: $$ \int_0^1 f(x^n)dx = \int_0^a f(x^n)dx +\int_a^1 f(x^n)dx $$ for some convenient $a=a(\varepsilon, f) \stackrel{\rm def}{= } 1-\frac{\varepsilon}{2\lVert f\rVert_\infty}$. The second part has absolute value at most $(1-a)\cdot\lVert f\rVert_\infty < \frac{\varepsilon}{2}$ by definition.

As for the first, we use continuity of $f$ at $0$: let $\delta_\varepsilon>0$ be such that $\lvert u\rvert \leq \delta_\varepsilon$ implies $\lvert f(u)\rvert \leq \frac{\varepsilon}{2}$. We then rely on the fact that for $n$ big enough (i.e., for $n\geq N_{\varepsilon}$, for some $N_{\varepsilon}\geq 0$, we have $\lvert x^n\rvert \leq \lvert a^n\rvert \leq \delta_\varepsilon$. Thus, the first term can be bounded as $$ \left\lvert \int_0^a f(x^n)dx\right\rvert \leq \int_0^a \left\lvert f(x^n)\right\rvert dx \leq \int_0^a \frac{\varepsilon}{2} dx \leq \frac{\varepsilon}{2} $$ for any $n\geq N_{\varepsilon}$.

Putting it together, for $n\geq N_{\varepsilon}$ one has $$ \left\lvert \int_0^1 f(x^n)dx\right\rvert \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$ by the triangle inequality. This shows that $$ \int_0^1 f(x^n)dx\xrightarrow[n\to\infty]{} 0 $$ and from there the general case follows.


Since $f $ is continuous, given $\varepsilon>0$ there exists $\delta>0$ such that $|f (x)-f (0)|<\varepsilon/2 $ whenever $|x|<\delta $. Let $M=\max \{|f (x)|:\ x\in [0,1]\}$. Choose $x_0\in (0,1) $ such that $1-x_0 <\varepsilon/2M $. Choose $n_0$ such that $|x_0^{n_0}|<\delta $ . Then, for any $n\geq n_0$,

\begin{align} \left|\int_0^1f (x^n)\,dx-f (0)\right| &=\left|\int_0^1(f (x^n)-f (0))\,dx\right| \leq \int_0^1|f (x^n)-f (0)|\,dx \\ \ \\ &= \int_0^{x_0}|f (x^n)-f (0)|\,dx + \int_{x_0}^1|f (x^n)-f (0)|\,dx \\ \ \\ &\leq \frac\varepsilon2 x_0+2M (1-x_0)\\ \ \\ &<\frac\varepsilon 2+\frac\varepsilon 2=\varepsilon. \end{align}

As an aside, if you have the Fundamental Theorem of Calculus available, the substitution $u=x^n $ does the job.

  • $\begingroup$ Isn't it exactly the same argument as in my answer? $\endgroup$
    – Clement C.
    Jan 17, 2017 at 1:11
  • $\begingroup$ very nice proof, professor as you made it much easier to follow as an undergrad student at a state university. $\endgroup$
    – DeepSea
    Jan 17, 2017 at 1:12

This question popped up recently (2022) and we were redirected here as it was flagged as a duplicate. So here is another two cents opinion. (Not worth more.)

As evident here it is difficult for advanced students not to suggest that this is immediate from the Lebesgue dominated convergence theorem. Indeed, while this is stated for continuous functions (with a plea to use the regulated integral) it is true in greater generality:

Theorem 1. Suppose that $f:[0,1]\to\mathbb R$ is Lebesgue integrable and that $f$ is continuous at $0$. Then $$ \left| \int_0^1 f(x^n)\,dx - f(0)\right| = \left| \int_0^1 [f(x^n)-f(0)]\,dx \right| \leq \int_0^1|f(x^n)-f(0)|\,dx \to 0 \tag{1} $$ as $n\to \infty$.

The proof (as most will rush to assert) follows from the fact that the sequence of functions $\{f(x^n)-f(0)\}$ converges a.e. to zero and is dominated by a Lebesgue integrable function. Hence Lebesgue's dominated convergence theorem applies.

Too advanced? Well yes for students restricted to the regulated integral or the archaic Riemann integral. But there is a well-known theorem that is not advanced, is not modern (it belongs in the 19th century) and it is adequate to prove this version.

Theorem 2. Suppose that $f:[0,1]\to\mathbb R$ is regulated. Then $$ \left| \int_0^1 f(x^n)\,dx - f(0+)\right| = \left| \int_0^1 [f(x^n)-f(0+)]\,dx \right| \leq \int_0^1|f(x^n)-f(0+)|\,dx \to 0 \tag{2} $$ as $n\to \infty$.

While we cannot appeal to Lebesgue's version of a convergence theorem there is an earlier weaker version known as the Arzelà-Osgood bounded convergence theorem. Osgood proved it for continuous functions and Arzelà for Riemann integrable functions.

Theorem [Arzelà-Osgood] Suppose that $\{f_n\}$ is a sequence of Riemann integrable functions on an interval $[a,b]$ that is (i) uniformly bounded and (ii) converges pointwise to another Riemann integrable function $f$. Then $ \lim_{n\rightarrow \infty}\int_a^b f_n(x)\,dx = \int_a^b f(x)\,dx.$

Proof of Theorem 2. The sequence $\{f(x^n)-f(0+)\}$ converges to zero pointwise on $(0,1)$ since $f$ is regulated. It is constant at $x=0$ and $x=1$. Thus we have it converging pointwise everywhere to another regulated function whose integral is certainly zero. Regulated functions are bounded and hence this sequence is uniformly bounded. Accordingly the Arzelà-Osgood completes the proof.

Didn't learn the Arzelà-Osgood theorem? Shame on your instructor. Shame on your textbook. As long as the powers that be have determined that students learn the Riemann integral or its more attractive cousin, the regulated integral, rather than the Lebesgue integral, they should teach the full theory.

  • 1
    $\begingroup$ After a few years of advancing beyond this point I agree with you. $\endgroup$ Feb 16, 2022 at 20:09

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