Exercise 7.4.10

Question

Assume $g$ is integrable on $[0,1]$ and continuous at 0 . Show
$$
\lim _{n \rightarrow \infty} \int_{0}^{1} g\left(x^{n}\right) d x=g(0)
$$

Ulisse Mini · Accepted Answer

Let $|g| < M$ and let $M > \epsilon > 0$. Define the partition $P = \{0, 1-\epsilon/(4M), 1\}$. Since $g$ is continuous, we can find $N$ so that for $n > N$, $$\left|g\left(\left(1- \frac{\epsilon}{4M}\right)^n\right) - g(0)\right| < \frac{\epsilon}{4}$$ Thus, $$\begin{aligned} U(g(x^n), P) &\leq \left(g(0) + \frac{\epsilon}{4}\right) \left(1-\frac{\epsilon}{4M}\right) + \frac{M\epsilon}{4M} \ &\leq g(0) + \frac{\epsilon}{4} + \left|\frac{2M \epsilon}{4M}\right| + \frac{\epsilon}{4} = g(0) + \epsilon \end{aligned} $$ A similar argument can be made that $L(g(x^n), P) \geq g(0) - \epsilon$.Assuming that $g(x^n)$ is integrable, this implies that $$\left|g(0) - \int^1_0 g(x^n) dx\right| < \epsilon$$ completing the proof.

Exercise 7.4.10

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Exercise 7.4.10

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