Proving g(x) is integrable, knowing f(x,y) is integrable

Question

I am a self learner on measure theory and am following the notation on Bartle, but when I was looking at the following problem I was not able to draw out a convincing approach and would like to know if somebody may have an idea as to what should be done. I know the other implication isn't true. What I was thinking on doing was trying to prove that if there were to exist a function f(x,y) which is not lebesgue integrable over such interval, then it must be that g is not lebesgue integrable over [0,1]. I didn't get anywhere with this and have thought about it some time, but haven't connected the pieces correctly. I am real grateful for any ideas given.

Let $g:[0,1]\rightarrow R$ be a Lebesgue measurable function such that the function $f(x,y)=2g(x)-3g(y)$ is Lebesgue integrable over $[0,1]$ x $[0,1]$. Prove that g is Lebesgue integrable over $[0,1]$.

Answer 1

$\int_0^{1} \int_0^{1} |2g(x)-3g(y)| \, dx \, dy<\infty$. This implies, by Fubini/Tonelli's Theorem that $ \int_0^{1} |2g(x)-3g(y)| \, dx <\infty$ for almost all $y$. Just fix one such $y$ and note that $ \int_0^{1} |2g(x)| \, dx \leq \int_0^{1} |2g(x)-3g(y)| \, dx +3|g(y)|$ [Lebesgue measurability of $g$ is not a problem. For each fixed $y$, $x\to f(x,y)$ is Lebesgue measurable].

Answer 2

Since $f$ is integrable, Fubini's theorem implies that the integral over $[0,1]\times [0,1]$ of $f$ exists and can be computed iteratively via \begin{align*} \int_0^1 \int_0^1 f(x,y) dxdy &= \int_0^1 \int_0^1 2g(x)-3g(y) dxdy \\ &= \int_0^1\left( \int_0^1 2g(x)dx-3g(y)\right)dy \\ &= 2\int_0^1 g(x)dx - 3\int_0^1 g(y)dy \\ &= -\int_0^1 g(x)dx. \end{align*} Hence if $g$ were not integrable, then this last expression would be undefined, and so $f$ would also not be integrable.