$$\int \sqrt{\tan x} dx$$
\begin{align}
&\int \sqrt{\tan x} dx\\
& = \int \sqrt{\frac{\sin x}{\cos x}}dx\\
& = \int \sqrt{\frac{(\cos x)'}{-\cos x}}dx\\
& = \int \sqrt{-\frac{f'}{f}}dx\\
& = \int \sqrt{-(\ln(f))'}dx\\
& = \int \sqrt{(\ln(\frac{1}{f}))'}dx\\
\text{Let }
u^2 & =(\ln(\frac{1}{f}))'\\
2udu & =(\ln(\frac{1}{f}))''dx=2udx\implies du=dx\\
& \int \sqrt{(\ln(\frac{1}{f}))'}dx\\
& = \int \sqrt{u^2} du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C \neq ANS\\
\end{align}
No, the proof doesn't work.
If $f_n \to f$ and $\mu(f_n=\infty) = 0$ for all $n$, this does not imply that $\mu(f=\infty)=0$. As a counterexample, consider $f_n = n \mathbf{1}_{[0,1)}$ for all $n$.
A correct argument could go something like this: suppose $A = \{f = \infty\}$ has positive measur...
