How do we prove $\int_I\int_x^1\frac{1}{t}f(t)\text{ dt}\text{ dx}=\int_If(x)\text{ dx}$ Let $f:\mathbb{R}\to\mathbb{R}$ be Borel-measurable and Lebesgue-integrable over $I:=(0,1)$. Further, let
$\;\;\;\;\;\;\;\;\;\;g : I\to \mathbb{R}\;,\;\;\; \displaystyle x \mapsto\int_x^1\frac{1}{t}f(t)\text{ dt}$
I want to show that $g$ is integrable over $I$ and that it holds
$\;\;\;\;\;\;\;\;\;\;\displaystyle\int_Ig(x)\text{ dx}=\int_If(x)\text{ dx}$
Proof: $\;\;\;$ Please note that $g$ is well-defined, because $f$ is Lebesgue-integrable over $I$. So, by Fubini's theorem $g$ is integrable and it holds:
$\;\;\;\;\;\;\;\;\;\;\displaystyle\int_Ig(x)\text{ dx}=\displaystyle\int_I\int_x^1\frac{1}{t}f(t)\text{ dt}\text{ dx}=\displaystyle\int_x^1 \int_I\frac{1}{t}f(t)\text{ dx}\text{ dt}=\displaystyle\int_x^1 \frac{1}{t}f(t)\text{ dt}=\cdots$
How do we integrate $\int_x^1 \frac{1}{t}f(t)\text{ dt}?$
 A: Fubini's Theorem says that since $f$ is integrable and for each $a\gt0$, $\frac1t$ is bounded on $[a,1]$
$$
\begin{align}
\int_a^1\int_x^1\frac1tf(t)\,\mathrm{d}t\,\mathrm{d}x
&=\int_a^1\int_a^t\frac1tf(t)\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_a^1\frac{t-a}tf(t)\,\mathrm{d}t\\
&=\int_a^1f(t)\,\mathrm{d}t-\int_a^1\frac atf(t)\,\mathrm{d}t
\end{align}
$$
Note that the region of integration is $\{(x,t):a\le x\le t\le1\}$. When changing the order of integration, the limits for each integral must be adjusted to represent this region.
Looking at $\frac atf(t)$ on $[a,1]$, we see that $\left|\frac atf(t)\right|\le\left|f(t)\right|$, and pointwise,
$$
\lim_{a\to0^+}\frac atf(t)=0
$$
thus, Dominated convergence says that
$$
\begin{align}
\lim_{a\to0^+}\int_a^1\int_x^1\frac1tf(t)\,\mathrm{d}t\,\mathrm{d}x
&=\lim_{a\to0^+}\left(\int_a^1f(t)\,\mathrm{d}t-\int_a^1\frac atf(t)\,\mathrm{d}t\right)\\
&=\int_0^1f(t)\,\mathrm{d}t-0
\end{align}
$$
A: Let $h(x,t) = {1 \over t} |f(t)| 1_{[x,1)}(t)$. Note that $g(x) = \int_I h(x,t) dt$.
Since $h(x,t) \ge 0$ for $x,t \in I$, Tonelli gives  $\int h = \int (\int h(x,t) dt) dx = \int( \int h(x,t) dx) dt$.
We have $\int( \int h(x,t) dx) = |f(t)|$, and so $\int( \int h(x,t) dx) dt = \int |f| < \infty$. Hence $g$ is integrable.
Since $h$ is integrable, the function $\eta(x,t) = {1 \over t} f(t) 1_{[x,1)}(t)$ is integrable, and $g(x) = \int \eta(x,t) dt$. Applying Fubini (as in Did's comment above) gives
$\int h = \int g = \int f$.
