Use Monotone Convergence Theorem to prove that
$$f(x) = [(2x-4)^{-1/2}]1_{(2,4]} + 01_{\{2\}}$$
is Lebesgue integrable on $[2,4]$.
According to notes from a different class
So I just replace $x$ w/ $f(x)$?
It seems then that I just say that MCT gives us:
$$\lim_{n \to \infty} \int_{[2,4]} g_n(x) dx = \int_{[2,4]} f(x) dx = \int_{[2,4]} |f(x)| dx$$
I just evaluate
$$\int_{[2,4]} g_n(x) dx$$
then take the limit. If it's $< \infty$, QED?
Changing variables you can work with the integral:
$$\int_0^1 f(x) \, dx=\int_0^1 x^{-1/2} \, dx$$
Consider the sequence of simple functions $\phi_n \leqslant f$ where
$$\phi_n(x) = \sum_{k=1}^{2^n}(k/2^n)^{-1/2}1_{[(k-1)/2^n, k/2^n]}(x).$$
Since the sequence converges monotonically to $f$ it follows by the MCT that,
$$\int_{[0,1]}f=\lim_{n \to \infty}\int\phi_n= \lim_{n \to \infty}\sum_{k=1}^{2^n}(k/2^n)^{-1/2}m([(k-1)/2^n,k/2^n])\\= \lim_{n \to \infty}\frac1{\sqrt{2^n}}\sum_{k=1}^{2^n}\frac1{\sqrt{k}}= \lim_{n \to \infty}\frac1{\sqrt{n}}\sum_{k=1}^{n}\frac1{\sqrt{k}} =2.$$