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}