$\textbf{Theorem:}$ Let $A,B$ two locally compact groups and $G=A\times B$ their product. Let $\mu_A$ and $\mu_B$ a left Haar measures. Then $\mu_A \times \mu_B$ is a left Haar measure of $G$.
$\textbf{Corollary:}$ Let $\bigtriangleup_A$ and $\bigtriangleup_B$ the modular functions of $A$ and $B$. Then
$$ \bigtriangleup_G(a,b)=\bigtriangleup_A(a)\bigtriangleup_B(b)$$
My proof:
Suppose that $A$ and $B$ are $\sigma-$finite then $\mu_A \times \mu_B$ satisfies:
$$ (\mu_A \times \mu_B) (M) = \int_A \mu_B (M_x) \mu_A(dx) $$
where $M$ is a borel set of $\mathcal{B}(A) \times \mathcal{B}(B)$ and $M_x = \{y \in B \hspace{0.5mm} \vert \hspace{0.5mm} (x,y) \in M \}$. It's easy to see that for $g=(g_1,g_2)\in G$ we have:
$$ (gM)_x = g_2 M_{g_1^{-1}x}$$
So:
$$ (\mu_A \times \mu_B) (gM) = \int_A \mu_B ((gM)_x) \mu_A(dx) = \int_A \mu_B (g_2 M_{g_1^{-1}x}) \mu_A(dx) = \int_A \mu_B (M_{g_1^{-1}x}) \mu_A(dx) $$
Then if put $y=g_1^{-1}x$ we have $\mu_A(dy)=\mu_A(dx)$ and:
$$ (\mu_A \times \mu_B )(gM) = \int_A \mu_B (M_{y}) \mu_A(dy) = (\mu_A \times \mu_B)(M)$$
The corollary is immediate.
My question is how to prove it for the case where it is not $\sigma$-finite. The theorem statement does not mention that $\mu_A$ and $\mu_B$ are $\sigma$-finite.
This is not true as written.
For example, if $A$ is a locally compact Hausdorff group of cardinality greater than $\mathfrak c$, then $A\times A$ is Hausdorff, so the diagonal is Borel. On the other hand, it is not in the product sigma algebra $\mathcal B(A)\otimes \mathcal B(A)$ (see e.g. here).
It follows that $\mu_A\otimes \mu_A$ is not a Borel measure on $A\times A$, so it is not the Haar measure, even though it is left invariant on the product algebra (which follows easily from the definition), as well as inner and outer regular.
It is true if $A,B$ are second-countable (for example, they are real or complex Lie groups with countably many connected components), however, since in this case $\mathcal B(A)\otimes \mathcal B(B)=\mathcal B(A\times B)$. For locally compact groups, $\sigma$-finiteness follows from second countability (in fact, I think the two are equivalent, but I did not check this).
The Haar theorem implies that in general, $\mu_A\otimes \mu_B$ can be uniquely extended to a Haar measure on $A\times B$, and so the corollary is also true with no hypotheses about $A$ and $B$.