Prove that $\lim_{n \to \infty} \int_{[0,1]}{x^n}\, dx = 0$. Where $\int$ represents Lebesgue integration.

Please check my proof, thank you.

Prove \begin{align*} \lim_{n \to \infty} \int_{[0,1]}{x^n}\, dx = 0 \end{align*} Proof. Let $$f_n(x) = x^n$$. Since $$f_n$$ is a polynomial it is continuous and therefore measurable. It is clear that $$(x^n) \to 0$$ for $$x \in [0, 1)$$ and $$(x)^n \to 1$$ for $$x = 1$$. Furthermore, $$0 \leq x^{n+1} \leq x^{n}$$ for all $$x \in [0,1]$$ and $$n\in \mathbb{N}$$. So, $$f_n$$ is measurable, nonnegative, and decreasing, and $$f_n \to 0$$ almost everywhere in $$[0,1]$$. By the monotone dominated convergence theorem \begin{align} \lim_{n \to \infty} \int_{[0,1]}{x^n}\, dx = \int_{[0,1]}{0}\, dx = 0 \cdot m{[0,1]} = 0 \end{align}

Edit. So apparently the monotone convergence theorem applies to only increasing sequences of functions. So for the problem above we could let $$g_n(x) = x - x^n$$. Now $$(g_n)$$ is an monotone increasing sequence that is nonnegative and measurable (continuous). Also $$(g_n) \to x$$ almost everywhere in $$[0,1]$$ (not at $$x = 1$$). So by the monotone convergence theorem

\begin{align} \lim_{n \to \infty} \int_{[0,1]}{g_n(x)}\, dx = \int_{[0,1]}{x}\, dx = \int_{[0,1]}\left(f_1(x) - 0\right) \, dx \end{align}

also we have

\begin{align} f_1(x) &= \left(x - x^k\right) + x^k \\ \\ &\implies \int_{[0,1]}x = \int_{[0,1]} \left(x - x^k\right) + \int_{[0,1]} x^k \\ \\ &\implies \left(\int_{[0,1]}x\right) - \left(\int_{[0,1]} x^k\right) = \int_{[0,1]} \left(x - x^k\right) \\ \\ &\implies \lim_{k \to \infty} \left[ \left(\int_{[0,1]}x\right) -\left(\int_{[0,1]} x^k\right) \right] = \lim_{k \to \infty} \int_{[0,1]} \left(x - x^k\right) = \lim_{k \to \infty} \int_{[0,1]} g_k(x) \\ \\ &\implies \left(\int_{[0,1]}x\right) - \lim_{k \to \infty} \left(\int_{[0,1]} x^k\right) = \left(\int_{[0,1]} x\right) \\ \\ &\implies - \lim_{k \to \infty} \int_{[0,1]} x^k = 0 \\ \\ &\implies \lim_{k \to \infty} \int_{[0,1]} x^k = 0 \end{align}

• I think MCT applies to monotone increasing sequences. But you can use DCT for the same result. – user25959 Nov 12 '18 at 2:07
• Why not just compute the integral and then take the limit? – user608030 Nov 12 '18 at 2:33
• How do you compute the lebesgue integral of $x^n$? – Zduff Nov 12 '18 at 2:39
• By noting that it's the same as the Riemann integral. Unless you don't have that theorem then just use dominated convergence like user25959 said. – user608030 Nov 12 '18 at 2:48
• @user25959 you can apply also MCT here without much effort, taking $g_n(x):=-f_n(x)+1$ and $g(x):=-f(x)+1$, and after applying elementary properties of the Lebesgue integral – Masacroso Nov 12 '18 at 3:08