Integral over product space with one singularity is finite Here's the question:

Let $f:(0,1) \to \mathbb R$ be a Lebesgue measurable function in $L^1((0,1))$. Define g on $(0,1) \times (0,1)$ by $g(x,y) = \frac{f(x)}x \chi(\{(x,y):y < x\}) $
a) Prove that $g$ is Lebesgue measurable
b) Prove that the function $g$ is integrable in $(0,1) \times (0,1)$. State (without proof) any theorems used in the proof.

I have proved a) by expressing $g^{-1}(a, \infty)$ as a measurable set.
The only relevant theorem I can think of using in b) is Foubini Tonelli, which allows us to compute or bound the integral as a nested one, and allows us to switch the order of integration. I can see that:
$\int_{(0,1) \times (0,1)}|g| d(\lambda \otimes \lambda) \leq \int_{(0,1)}\frac1y\big(\int_{(y,1)}|f(x)|d\lambda(x)\big)d\lambda(y)$,
but I don't see how I can get a bound that is actually finite.

I realised, one can compute the integral the other way around:
$\int_{(0,1) \times (0,1)}|g|d(\lambda \otimes \lambda) = \int_{(0,1)}\int_{(0,x)}|f(x)/x|d\lambda(y)d\lambda(x) = \int_{(0,1)}|f(x)|d\lambda(x) < \infty$
