Lebesgue integral and limit; $\lim_{a\to 0^+}\int_{a}^{1}(t\ln(t))^3dt=\int_{0}^{1}(t\ln(t))^3dt$

Question

How can I prove:

$$\lim_{a\to 0^+}\int_{a}^{1}(t\ln(t))^3dt=\int_{0}^{1}(t\ln(t))^3dt$$

assuming that the integral are Lebesgue integrals. There may be a theorem that confirms this equality.

Answer 1

A more elementary approach could be the following. The function $t\mapsto t\ln t$ is continuous and bounded in $(0,1]$, just notice that $\lim_{t \to 0^+}t\ln t=0$, therefore if we set $$ f(t):=\begin{cases} t\ln t,& t>0\\ 0,& t=0 \end{cases} $$

then $f$ is continuous in $[0,1]$, so $f^3$ is Riemann integrable, so the improper integral coincides in value with the Riemann integral. Now, every integral in these expressions is trivially Lebesgue integrable as $f$ is continuous and bounded, so we are done.