Is $I$ Lebesgue integrable? 
Let 
  $$I=\int_0^\infty\frac{\ln(x)}{(1+x^2)\sqrt{x}}f(x)\,dx$$
  where $f(x)$ is a discontinuous function that equals $1$ a.e. on $[0, \infty)$.
Prove $I$ exists as a Lebesgue integral.

Suppose
$$f_n(x)=\begin{cases}0,\,\,\,\,\,&x=1/n\\1,&\text{otherwise}.\end{cases}$$
Here the set $\{1/n\}, n\in \mathbb{N}$ has measure $0$ and so $f_n=1$ a.e. on $[0,\infty).$ Also $f_n\leq f_{n+1}$, and $f_n\to 1$ a.e.
By MCT $$I=\int_0^\infty\frac{\ln(x)}{(1+x^2)\sqrt{x}}\,dx$$
by LCT with $1/(1+x^2)$ $I$ convergese and thus is Lebesgue integrable.
I am not sure whether this is correct. If so, I proved this for only a specific $f(X)$, how would one go about the general proof?
 A: Perhaps do it in this way:
\begin{align*}
\int_{0}^{\infty}\dfrac{|\log x|}{(x^{2}+1)\sqrt{x}}|f(x)|dx&=\int_{0}^{\infty}\dfrac{|\log x|}{(x^{2}+1)\sqrt{x}}dx=\int_{0}^{1}\dfrac{\log(1/x)}{(x^{2}+1)\sqrt{x}}dx+\int_{1}^{\infty}\dfrac{\log x}{(1+x^{2})\sqrt{x}},
\end{align*}
where
\begin{align*}
\int_{0}^{1}\dfrac{\log(1/x)}{(x^{2}+1)\sqrt{x}}dx\leq\int_{0}^{1}\dfrac{\log(1/x)}{\sqrt{x}}dx\leq C\int_{0}^{1}\dfrac{1}{x^{1/4}}\dfrac{1}{x^{1/2}}dx<\infty,
\end{align*}
and
\begin{align*}
\int_{1}^{\infty}\dfrac{\log x}{(1+x^{2})\sqrt{x}}\leq C\int_{1}^{\infty}\dfrac{\sqrt{x}}{(1+x^{2})\sqrt{x}}dx=\int_{1}^{\infty}\dfrac{1}{1+x^{2}}dx<\infty.
\end{align*}
A: You do not know how complicated the set $N$ on which the values of $f$ are not specified but you do know that it has measure zero. Therefore, 
$\int_0^\infty\left|\frac{\ln(x)}{(1+x^2)\sqrt{x}}f(x)\right|\,dx=\int_{\mathbb R^{\ge0}\setminus N}\frac{|\ln(x)|}{(1+x^2)\sqrt{x}}\,dx\le \int^{\infty}_0\frac{|\ln(x)|}{(1+x^2)\sqrt{x}}\,dx=-\int^1_0\frac{\ln(x)}{(1+x^2)\sqrt{x}}dx+\int^{\infty}_1\frac{\ln(x)}{(1+x^2)\sqrt{x}}dx.$
Now use elementary calculus to show that each of these integrals converges. For the first one note that $\int\frac{\ln\left(x\right)}{\sqrt{x}}dx=2\sqrt{x}\ln \left(x\right)-4x^{\frac{1}{2}}+C$ and for the second one, note that $\frac{\ln\left(x\right)}{\sqrt{x}}$ is bounded on $[1,\infty)$.
