If $f$ is Lebesgue integrable on [0,1] show $g(x)=\int_{[x,1]} f(t)t^{-1}dt$ is Lebesgue integrable on [0,1]

Question

Also want to show $\int_{[0,1]}g(x)dx = \int_{[0,1]}f(x)dx$.

So since $f \in \mathcal{L}([0,1]), f=u-v$ where $u$ and $v$ are upper functions. Then I need to show $\int_{(x,1]} u(t)t^{-1}dt$ is an upper function because that will give $g \in \mathcal{L}([0,1])$.

Not really sure how to go about doing this. I have to show this satisfies the properties of being an upper function but that's what I'm having trouble starting.

Answer 1

$\int_0^1|g(x)|dx \leq \int_0^1 dx\int_x^1 \left|\dfrac{f(t)}{t}\right|dt = \int_0^1 dt \int_0^t \left|\dfrac{f(t)}{t}\right|dx = \int_0^1 \left|f(t)\right|dt$