Assume $f(x)\in L^1(0,1)$, prove that $g(x) = \frac{1}{x}\int_{0}^{x}\frac{f(t)}{log(t)}dt$ is in $L^1(0,1)$ Assume $f(x)\in L^1(0,1)$, prove that $g(x) = \frac{1}{x}\int_{0}^{x}\frac{f(t)}{log(t)}dt$ is in $L^1(0,1)$
Have a hard time knowing where to start...
 A: \begin{align}
\int_0^1 g(x)dx &= \int_0^1\int_0^x\frac{1}{x}\frac{f(t)}{\log t}dt dx \\
&= \int_0^1\int_t^1\frac{1}{x}\frac{f(t)}{\log t}dx dt \\
&= \int_0^1\int_t^1\frac{dx}{x} \cdot\frac{f(t)}{\log t}dt \\
&= \int_0^1\log t \cdot\frac{f(t)}{\log t}dt \\
&= \int_0^1 f(t)dt
\end{align}
A: For first,
$$ g(x)=\int_{0}^{1}\frac{f(xu)}{\log x +\log u}\,du =\int_{0}^{1}\frac{f}{\log}(xu)\,du.\tag{1}$$
For any $t\in(0,1)$, let $\mu(t)$ be the length of the curve whose support is given by:
$$ \{(x,u)\in(0,1)^2:xu=t\}.\tag{2} $$
In order to prove that $g\in L^1(0,1)$, we just need to prove that 
$$ w(t)=\frac{\mu(t)}{\left|\,\log t\,\right|} \tag{3}$$
is bounded on $(0,1)$. Since:
$$\mu(t)=\int_{t}^{1}\sqrt{1+\frac{t^2}{x^2}}\,dx\leq\int_{t}^{1}\left(1+\frac{t}{x}\right)\,dx = t-1+t\log t\tag{4}$$
it follows that $w(t)\leq 2$, hence:

$$ \|g\|_{L^1}\leq 2\cdot\|f\|_{L^1}.\tag{5} $$

Once we know that $g\in L^1$, we are allowed to switch the order of integration, and Fubini's theorem gives $\|g\|_{L^1}=\|f\|_{L^1}$ as shown in the other answers. 
A: Methodology hints
Write out the (double) integral you want to show is bounded. Then flip the order of integration and you'll be able to perform the $x$ integration. You'll be left with an integral extremely similar to the $L^1$ condition for $f$.
