In my continuing studies towards an upcoming exam, I have run across a number of problems similar to the following, and would appreciate 1) some guidance on how to approach such problems in general and 2) how to solve this one in particular.
Let $f \in L^1[0,1]$ and define $g(x) = \int_{[x,1]} \frac{f(t)}{t} dt$. Show that $g \in L^1[0,1]$ and $\int_{[0,1]} g = \int_{[0,1]} f$.
I suspect the use of Tonelli's theorem is required, so I wrote
\begin{align*} \int_{[0,1]} |g| &= \int_{[0,1]} \left| \int_{[x,1]} \frac{f(t)}{t} dt \right| dx \\ &= \int_{[0,1]} \left| \int_{[0,1]} \chi_{[x,1]} \frac{f(t)}{t} dt \right| dx \\ &= \int_{[0,1]} \left| \frac{f(t)}{t} \int_{[0,1]} \chi_{[x,1]} dx \right| dt \end{align*}
However, I'm not sure if this is even a productive direction. Thanks for any assistance.
Since it should hold for all $f \in L^1([0,\,1])$, it would in particular hold for positive $f$, so you must be able to move the absolute value into the inner integral. That yields
$$\begin{align} \int_{[0,1]} \lvert g(x)\rvert\,dx &= \int_{[0,1]}\left\lvert \int_{[x,1]} \frac{f(t)}{t}\,dt\right\rvert\, dx\\ &\leqslant \int_0^1 \int_x^1 \frac{\lvert f(t)\rvert}{t}\,dt\,dx\\ &= \int_0^1\int_0^t \frac{\lvert f(t)\rvert}{t}\,dx\,dt\\ &= \int_0^1 \left(\int_0^t dx\right)\frac{\lvert f(t)\rvert}{t}\,dt\\ &= \int_0^1 \lvert f(t)\rvert\, dt. \end{align}$$
Changing the order of integration is no problem because the integrand is measurable and non-negative.
The same computation without the absolute value yields (we now know that Fubini's theorem can be applied)
$$\int_0^1 g(x)\,dx = \int_0^1 f(t)\,dt.$$