Finiteness In Tonelli's And Fubini's Theorems Why do we suppose in Tonelli's Theorem that  $(X,Σ_X , μ_x)$ and  $(Y,Σ_Y , μ_y)$ to be σ-finite? Does the theorem fail when it is not? If yes, could someone provide an example?
We also suppose in Fubini's theorem that : 
$\int_{X×Y}|f(x,y)|$ $d(μ_x×μ_y) < \infty$
Does that mean it is not true when the integral = $\infty$ ? If yes, could someone provide an exmaple?

Indeed Tonelli's theorem can fail if the spaces are not $\sigma$-finite. Take for instance $\mu_x$ the Lebesgue measure and $\mu_y$ to be the counting measure on $[0,1]$. Take $E=\{(x,x): x \in [0,1]\}$. Then $E$ is $\mathscr{B}(\mathbb{R})\otimes \mathscr{B}(\mathbb{R})$ measurable (it is closed) and
 $$\int_0^1\int_0^1 1_{E}(x,y) \, d\mu_x \, d\mu_y = \int_0^1 0 \, d\mu_y =0$$
and
$$\int_0^1\int_0^1 1_{E}(x,y) \, d\mu_y \, d\mu_x =
 \int_0^1 1\, d\mu_x = 1\neq 0.
$$
Here we had all the conditions for Tonelli's theorem except $\sigma$-finiteness, even one of the two spaces was a finite measure, but we could not exchange the integrals.
For Fubini's theorem, you don't actually need $f \in L^1$. Actually $f^+ \in L^1$ or $f^- \in L^1$ is enough. The proof is just apply Tonelli to the positive and negative parts of $f$. If we don't have either of $f^+,f^- \in L^1$, then it's not that the theorem isn't true, but more that the integrals aren't even defined. That is, it would make no sense to assert that two undefined quantities are equal. We can say that if either side is defined, then it must be that $f^+ \in L^1$ or $f^- \in L^1$, in which case both sides are defined and equal.
Finally, we can do a little better than $\sigma$-finiteness. A generalization that doesn't make the proof any more difficult is that $\mu_x,\mu_y$ can be $\Sigma$-finite, where $\mu$ is called $\Sigma$-finite if
$$
\mu = \sum_{n=1}^\infty \mu_i
$$
for some finite measures $\mu_i$. Note that every $\sigma$-finite measure is $\Sigma$-finite by taking $\mu_i(A) := \mu(A \cap E_i)$ where the $E_i$ are a partition of the whole space with $\mu(E_i) <\infty$ for all $i$.
