Let $f \in L^1([0,1])$. Define for $x \in [0,1]$: $g(x):= \int_x^1 t^{-1}f(t) d\mathcal{L}^1(t)$.
I want to show that $g \in L^1([0,1])$ and $\int_0^1 g(x) d\mathcal{L}^1(x)=\int_0^1 f(t) d\mathcal{L}^1(t)$.
Assuming that $g$ is integrable, one can apply Fubini to get:
$\int_0^1 g(x) d\mathcal{L}^1(x) = \int_0^1 \Big{(} \int_x^1 t^{-1}f(t) d\mathcal{L}^1(t) \Big{)} d\mathcal{L}^1(x) \\ =\int_0^1 \Big{(}\int_0^1 \chi_{[x,1]}(t) t^{-1}f(t) d\mathcal{L}^1(t) \Big{)} d\mathcal{L}^1(x) \\ =\int_0^1 \Big{(} \int_0^1 \chi_{[x,1]}(t) t^{-1}f(t) d\mathcal{L}^1(x)\Big{)} d\mathcal{L}^1(t) \\ =\int_0^1 \Big{(} \int_0^1 \chi_{[x,1]}(t) d\mathcal{L}^1(x) \Big{)} t^{-1}f(t) d\mathcal{L}^1(t)\\ =\int_0^1 f(t) d\mathcal{L}^1(t)$
since $\int_0^1 \chi_{[x,1]}(t) d\mathcal{L}^1(x)=t$.
So my question is how do I show that $\int_0^1 \Big{|} \int_x^1 t^{-1}f(t) d\mathcal{L}^1(t) \Big{|}d\mathcal{L}^1(x) < \infty$ ?