Given the Lebesgue measure $m_2$ on $(\mathbb{R}^2,\mathbb{B}_2)$, I have the following integral: $$\mu(\mathbb{R}^2)=...=\int 1_{[1,\infty )}\frac{1}{x^2}dm(x)$$ My professor insist on using the Monotone Convergence Theorem: $$\mu(\mathbb{R}^2)=\int 1_{[1,\infty )}\frac{1}{x^2}dm(x)=\lim_{n\to\infty}\int 1_{[1,n]}\frac{1}{x^2}dm(x)=\lim_{n\to\infty}\int_{1}^n\frac{1}{x^2}dx$$ $$=\lim_{n\to\infty}\left(-\frac{1}{n}-\left(-\frac{1}{1}\right)\right)=1$$ Why can I not just make it into the Riemann-integral from first step as: $$\mu(\mathbb{R}^2)=\int 1_{[1,\infty )}\frac{1}{x^2}dm(x)=\int_{1}^\infty\frac{1}{x^2}dx=...,$$ instead of using the Monotone Convergence Theorem? Because the requirements, that $\frac{1}{x^2}$ should be positive and continuous are fulfilled.
Any help would be greatly appreciated, as I cannot se why this step is needed.