# Limit of integrals of continuous functions

If $$g:[0,1] \rightarrow \mathbb{R}$$ be a continuous function such that $$\lim\limits_{x \rightarrow 0^+}\dfrac{g(x)}{x}$$ exist and is finite, then prove that

$$\lim\limits_{n \rightarrow \infty} \int_{0}^{1}g(x^n)dx= \int_{0}^{1} \frac{g(x)}{x}dx$$

This is problem from the book PROBLEMS IN REAL ANALYSIS- ADVANCE CALCULUS ON THE REAL AXIS by Titu Andreescu. The solution given on the book can't be understood by me. Please provide any alternative solution.

• Please include the hypothesis that $g$ is continuous in the statement of the problem (putting it vaguely in the title isn't good enough) ... Sep 11, 2019 at 17:01

This is false. Consider $$g(x) = x$$, then $$\int_0^1 g(x^n)dx = \int_0^1 x^ndx = \frac{x^{n+1}}{n+1}|_{x=0}^{x=1} = \frac{1}{n+1} \rightarrow 0$$ $$\int_0^1 \frac{g(x)}{x}dx = \int_0^1 dx = 1$$

Edit: It is true however that $$\lim_{n\rightarrow \infty} \int_0^1 n g(x^n)dx = \int_0^1 \frac{g(x)}{x}dx$$

If you change variables you find $$\int_0^1 n g(x^n)dx = \int_0^1 y^{1/n} \frac{g(y)}{y}dy$$ and apply dominated convergence as $$y^{1/n} \frac{g(y)}{y} \rightarrow \frac{g(y)}{y} \text{ for } y\in(0,1]$$ which is true from our assumption on $$g$$.

• Yes You are right..But the strange thing is this is proved in the book..and this is a reputed book..my goodness.. Could we trust anyone..it is really hurting.. Sep 11, 2019 at 16:43
• Can you provide the proof? It's possibly a typo on the question. Sep 11, 2019 at 16:44
• Yes I can. Yeah I also think that there is a typo in the question. Sep 11, 2019 at 16:54
• Perhaps my edit fixes the mistake? Sep 11, 2019 at 16:59
• Right, they just forgot the $n$ in the statement. No worries! Sep 11, 2019 at 18:26

Define $$h:[0,1] \rightarrow \mathbb{R}$$ by

$$h(t) = \begin {cases} g(t)/t , & t \in (0,1] \\ lim_{x \rightarrow 0^+}~g(x)/x, & t = 0 \end{cases}$$

Then $$h$$ is continuous and we can set $$H(x) = \int_0^1 h(t)dt.$$

We have

$$n \int_0^1g(x^n)dx=n \int_0^1x^nh(x^n)dx=xH(x^n)|_0^1- \int_0^1H(x^n)dx=H(1) - \int_0^1H(x^n)dx= \int_0^1 \frac{g(x)}{x}dx - \int_0^1H(x^n)dx.$$

If $$0 < a<1$$, then

$$| \int_0^1H(x^n)dx| \leq \int_0^1|H(x^n)|dx = \int_0^a|H(x^n)|dx+ \int_a^1|H(x^n)|dx \leq a|H( \alpha _n^n)| + (1-a)M~~~~~~~(1)$$

where $$\alpha _n^n \in [0,a]$$ and $$M=max_{t \in [0,1]}|H(t)|.$$

Consider $$\epsilon > 0$$ such that $$a>1 - \frac{\epsilon}{2M}$$. Since $$lim_{n \rightarrow \infty}|H( \alpha _n^n)|=0$$ it follows that $$a|H( \alpha _n^n)|< \epsilon /2$$ for all positive itegers $$n \leq N(\epsilon)$$. Relation (1) yields

$$| \int_0^1H(x^n)dx| \leq \epsilon /2 + (1-a)M < \epsilon /2 + (1-1+ \epsilon /2M)M= \epsilon$$

Hence, $$lim_{n \rightarrow \infty} \int_0^1 H(x^n)dx= 0$$ and the conclusion follows.