Can we conclude that $\int_0^1 f(x)dx$ is bounded if $\int_0^1 f(x)dx - \int_0^1 f(y)dy$ is bounded?
I have a question that asks if $f$ is measurable on $(0,1)$, and $f(x)-f(y)$ is integrable over the square $I\times I$, where $I=[0,1]$, then $f\in L(0,1)$. My attempt:
\begin{align*} -\infty<\int\int_{I\times I}(f(x)-f(y))dxdy &= \int_0^1 \left[\int_0^1(f(x)-f(y))dx\right]dy\\ &=\int_0^1 \left[\int_0^1f(x)dx-f(y)\right]dy\\ &=\int_0^1 \left[\int_0^1f(x)dx\right]dy-\int_0^1f(y)dy\\ &=\left[\int_0^1f(x)dx\right]\int_0^1 1dy-\int_0^1f(y)dy\\ &= \int_0^1f(x)dx-\int_0^1f(y)dy\\ &<+\infty, \end{align*} where we use Fubini's theorem on the fisrt equality. Can we say that $\int_0^1f(x)dx \in L(0,1)$?
By Fubini's theorem, the "iterated" integrals exist almost everywhere. In particular, $$\int_0^1(f(x)-f(y))\,dx=\int_0^1 f(x)\,dx-f(y)$$ exists for almost all $y$. Necessarily there exists $y$ for which $f(y)$ is finite since otherwise either $$f(x)-f(y)=\infty-\infty,$$ which is not defined, or $$f(x)-f(y)=f(x)-\infty$$ with $f(x)=-\infty$ but this wouldn't be integrable for any $x$ contrarily to what is above. Therefore, $f(y)$ is finite for some $y$ and so the same happens to $$\int_0^1 f(x)\,dx=\left(\int_0^1 f(x)\,dx-f(y)\right)+f(y).$$