Proving that a function is L1

Question

Suppose $f \in L^1([0,b])$ and $g(x)=\int_x^b{\frac{f(t)}{t}dt}$ , prove that $g\in L^1([0,b])$ and $\int_{0}^{b} g(x) dx = \int_{0}^{b} f(t) dt$. Assume we are not allowed to use integration by parts.

To say $g\in L^1([0,b])$, we have to show that the following integral is bounded:

$\int_{0}^{b} |(\int_{x}^{b} \frac{f(t)}{t}dt)|dx \leq \int_{0}^{b} (\int_{x}^{b} |\frac{f(t)}{t}|dt)dx$

I don't know how to continue since we have a singularity at $x=0$.

Answer 1

Hint: By Fubini's theorem, $$\begin{align} \int_0^b dx \int_x^b dt \left\lvert\frac{f(t)}{t}\right\rvert &= \int_0^b dx \int_0^b dt \mathbb{1}_{\{x\leq t\}}\left\lvert\frac{f(t)}{t}\right\rvert = \int_0^b dt \int_0^b dx \mathbb{1}_{\{x\leq t\}}\left\lvert\frac{f(t)}{t}\right\rvert \end{align}$$ For more:

$$\begin{align}\int_0^b dt \int_0^t dx \left\lvert\frac{f(t)}{t}\right\rvert = \int_0^b dt \left\lvert\frac{f(t)}{t}\right\rvert \int_0^t dx = \int_0^b dt\left\lvert\frac{f(t)}{t}\right\rvert\cdot t = \int_0^b dt \left\lvert f(t)\right\rvert \end{align}$$