decomposition of Haar measure and Fubini theorem

Question

Let $G$ have a unique decomposition $G=AB$, where $G,A,B$ are linear Lie groups with $G,A$ unimodular. Suppose we have Haar measure decomposition $dg=dad_rb$ where $d_rb$ is the right Haar measure on $B$, that is, we have

$$\int f(g)dg=\int f(ab) dad_rb$$

As far as I understand, $f$ has to be integrable over $G$, so that we apply Fubini theorem and get the above formula, in which case we can interchange the order of integration.

Now I'm curious how much the above formula depends on Fubini theorem?

Or, if we don't know $f$ is integrable, but we do have the right side iterated integral convergent. Moreover, we may assume $f$ is smooth, the function $\int f(ab)da$ is absolutely integrable as a function of $b$, do we still have the above formula?

Thank you very much for any of your comment, and sorry for the question if not appropriate here.

Answer 1

1. Yes, for applying Fubini you need $f \in L^1(G)$. You could also apply Tonelli though.

2. Smoothness is not at all an issue. If you assume that $b \mapsto \int f(ab) da$ is $L^1(B)$, then you can not conclude that $f \in L^1(G)$. Start with $G = \mathbb{R}^2$ for a counterexample. If you assume $f \geq 0$ additionally, you can conclude this.

Edit: See here for a circle: Iterated Integrals - "Counterexample" to Fubini's Theorem