Folland's Real Analysis: Modern Techniques and Their Applications states the Fubini-Tonelli Theorem as
Suppose that $(X, \mathcal{M},\mu)$ and $(Y, \mathcal{N},\nu)$ are $\sigma$-finite measure spaces.
a. (Tonelli) If $f \in L^+(X \times Y)$, then the functions $g(x) = \int f_x d\nu$ and $h(y) = \int f^y d\mu$ are in $L^+(X)$ and $L^+(Y)$, respectively, and $$ \int f d(\mu \times \nu) = \int \left[ \int f(x,y) d\nu(y) \right] d\mu(x) = \int \left[ \int f(x,y) d\mu(x) \right] d\nu(y).$$ b. (Fubini) If $f \in L^1(\mu \times \nu)$, then $f_x \in L^1(\nu)$ for a.e. $x \in X, f^y \in L^1(\mu)$ for a.e. $y \in Y$, the a.e.-defined functions $g(x) = \int f_x d\nu$ and $h(x) = \int f^y d\nu$ are in $L^1(\mu)$ and $L^1(\nu)$, respectively, and the above equation holds.
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So completeness is not necessary. On the other hand, $\sigma$-finiteness is necessary. This exercise follows the Fubini-Tonelli theorem in Folland. Here $\mathcal{B}_{[0,1]}$ denotes the usual Borel $\sigma$-algebra on $[0,1]$.
46. Let $X = Y = [0,1]$, $\mathcal{M} = \mathcal{N} = \mathcal{B}_{[0,1]}$, $\mu = $ Lebesgue measure, and $\nu =$ counting measure. If $D = \{(x,x): x\in [0,1]\}$ is the diagonal in $X \times Y$, then $\iint \chi_D \; d\mu\; d\nu$, $\iint \chi_D \; d\nu \; d\mu$, and $\int \chi_D \; d(\mu \times \nu)$ are all unequal.
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