Measure Product Theorem: may non-$\sigma$-finiteness result unique product?

Let $i\in\{1,2\}$. The Measure Product Theorem states that, given the measure spaces $(X_i,\Sigma_i,\mu_i)$, there is at least one product measure $\pi$ such that $\pi(A_1\times A_2)=\mu_1(A_1)\;\mu_2(A_2)$, for $A_i\in\Sigma_i$.

It also states that if the $\mu_i$'s are $\sigma$-finite, then $\pi$ is unique.

I'd like an example of a non-$\sigma$-finite pair of measures from which, nevertheless, we only obtain one product, thus showing "if" cannot be "iff" on the paragraph above.

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@DavideGiraudo Hmmm, you're right. I had actually found that question before, but not read it. I thought it was a "counter-counter-example", but it's about a counter. I will edit my question and ask about a double counter instead. – Luke Nov 20 '11 at 19:34
@All: This is no longer a duplicate. – t.b. Nov 20 '11 at 19:50
@t.b. You're right, I voted to close, but now I can't remove my vote. – Davide Giraudo Nov 20 '11 at 19:55
@Davide: You can't (I voted to close, too, before the question was changed). You could remove your "possible duplicate" comment, though (in the hope that people no longer vote for closure). – t.b. Nov 20 '11 at 20:00

Trivial example: $X = \{x\}$ has one point, $\mu(\{x\}) = \infty$, $X_1 = X_2 = X$, and $\mu_1 = \mu_2 = \mu$. Not $\sigma$-finite, but the unique product measure is $\lambda(\{(x,x)\}) = \infty$.
Take an uncountable set $X$ with counting measure defined on the power set and consider $X \times X$. Then the usual product measure and the complete locally determined product measure coincide with counting measure on the product and we conclude by the result that any measure $\lambda$ on the product $\sigma$-algebra satisfying $\lambda(A \times B) = \mu(A) \nu(B)$ for all sets $A$ and $B$ of finite measure must lie between the complete locally determined product and the usual product, as was explained in this answer.