Let $(X,\mathcal{M},\mu), (Y, \mathcal{N},\nu)$ be measure spaces. It's a standard result that if $f \in L^p(\mu), g \in L^q(\mu)$, and $p^{-1} + q^{-1} = 1$, then $$\int_X |fg| d \mu = \Vert fg \Vert_1 \le \Vert f \Vert_p \Vert g \Vert_q = (\int_X |f|^p d \mu)^{1/p} (\int_X |g|^q d \mu)^{1/q}$$
However, what if $f \in L^p (\mu)$ and $g \in L^q(\nu)$? Does Holder's inequality still hold?
The reason I am asking this question is that in a problem I am working on, I have $f \in L^2(\mu)$ and $g \in L^2(\nu)$, and I need to justify breaking the integral $\int_{X \times Y} (f(x)g(y)) d(\mu \times \nu) (x,y)$ up into individual integrals using Fubini-Tonelli, which only works if the product is in $L^1(\mu \times \nu)$.
Any help is appreciated!
If $\mu$ and $\nu$ are finite measures, then we always have also $f \in L^1(\mu)$ and $g \in L^1(\nu)$ because of the Hölder-Inequality (resp. Cauchy-Schwarz) and the integrability of the constant function $x \mapsto 1$.
Otherwise the corresponding Lebesgue integral is not definied. (For example take $\mu = \nu = \lambda$ and $f=g$, where $f$ is a function which is in $L^2$ but not in $L^1$.)
Note that, in order to use the Theorem of Fubini-Tonelli, you need $\sigma$-finite measures.