I want to calculate $\displaystyle\int_0^1 \dfrac{x}{1-x}(-\log x)\ dx$ in Lebesgue integral theory. I calculated this using the theorem below.
Theorem
Let $\{f_n \}_{n=1}^{\infty}$ be a sequence of functions that are Lebesgue-measurable and non negative. And let $f:=\sum_{n=1}^{\infty} f_n$.
Then, $\displaystyle\int f dx=\sum_{n=1}^{\infty} \displaystyle\int f_n (x) dx.$
My calculation
\begin{align}
\displaystyle\int_0^1 \dfrac{x}{1-x}(-\log x) \ dx
&=\displaystyle\int \dfrac{x}{1-x}(-\log x) \cdot \chi_{(0,1)} (x) \ dx\\
&=\displaystyle\int \bigg(\sum_{k=1}^{\infty} x^k\bigg)\cdot (-\log x) \cdot \chi_{(0,1)} (x) \ dx\\
&=\displaystyle\int \sum_{k=1}^{\infty} \bigg(x^k (-\log x) \cdot \chi_{(0,1)} (x)\bigg) \ dx\\
&=\sum_{k=1}^{\infty} \displaystyle\int \bigg(x^k (-\log x) \cdot \chi_{(0,1)} (x)\bigg) \ dx \ (\text{ using the Theorem) }\\
&=\sum_{k=1}^{\infty} \displaystyle\int_0^1 x^k (-\log x) \ dx \\
&=\sum_{k=1}^{\infty} \Bigg(\bigg[\dfrac{x^{k+1}}{k+1} (-\log x)\bigg]_0^1+ \displaystyle\int_0^1 \dfrac{x^{k}}{k+1} \ dx\Bigg) \ (\text{Integration by parts})\\
&=\sum_{k=1}^{\infty} \Bigg(0+ \dfrac{1}{(k+1)^2}\Bigg)
=\sum_{k=1}^{\infty}\dfrac{1}{(k+1)^2}.
\end{align}
But I wonder if I can change $\sum$ and $\displaystyle\int.$
For each $k$, $x^k(-\log x)\chi_{(0,1)}(x)$ is non negative but is it Lebesgue-measurable? And why?
Best Answer
You used the theorem correctly. Continuous functions are measurable, open sets are measurable and so characteristic functions of open sets are measurable, and products of measurable functions are measurable.