Computing the Lebesgue Integral of $\int_{[1, \infty)} \frac{1}{x^2} \; d\mu$. I am trying to compute the Lebesgue integral of $\int_{[1, \infty)}  \frac{1}{x^2} \; d\mu$. I don't think MCT can be used here. Based on the answers to this similar question, I thought that I should use DCT, but I am not sure how to find a suitable function that bounds the sequence of functions that converges pointwise to $\frac{1}{x^2}$.
 A: The function you try to integrate is continuous and positive so your Lebesgue integral is the same thing as an improper Riemann integral.
So : $\int_1^{+\infty} 1/x^2 dx=[-1/x]_1^{\infty}=1$
A: Consider the sequence $f_{n}(x)=\frac{1}{x^{2}}\chi_{[1,n]}$, where $\chi_{[1,n]}$ is the indicator function of $[1,n]$. This sequence satisfies the hypotheses of the DCT, and you can compute the integrals $\int f_{n}$ using Riemann integration. You therefore get that $\int_{[1,\infty)}\frac{1}{x^{2}}dx=\lim \int f_{n}$.
Let's check the hypotheses:
First, $f_{n}$ is measurable since the function $\frac{1}{x}$ is continuous in $(0,\infty)$ (hence measurable) and $\chi_{[0,1]}$ is measurable (since $[0,1]$ is too). Therefore $f_{n}$ is measurable as the product of two measurable functions.
Second, let $x\in [1,\infty)$. take $N\in \mathbb{N}$ to be the least integer greater than or equal to $x$. The sequence $(f_{n}(x))_{n\in \mathbb{N}}$ is given by: $f_{n}(x)=0$ if $n< N$, $f_{n}(x)=f(x)$ if $n\geq N$. This sequence is increasing and convergent to $f(x)$, as desired.
Hope this helps!
