In measure theoretic probability we have been taught the following version of Caratheodory's extension theorem:
Theorem: $1.$ Suppose $\mathcal S\subset \mathcal P(\Omega)$ is a semi-algebra(semi-field) of subsets of $\Omega$.If $\nu:\mathcal S\to [0,\infty]$ is a pre-measure i.e. $\nu(\phi)=0$ and $\nu$ is a $\sigma$-additive set function.Then there exists a $\sigma$-additive set function $\mu:\mathcal A(\mathcal S)\to [0,\infty]$ such that $\mu(A)=\nu(A)$ for all $A\in \mathcal S$.(Where $\mathcal A(\mathcal S)$ denotes the smallest algebra containing $\mathcal S$)
$2.$ If $\mathcal A\subset \mathcal P(\Omega)$ is an algebra(or field) on $\Omega$ and $\mu:\mathcal A\to [0,\infty]$ is a $\sigma$-additive set function with $\mu(\phi)=0$,then there exists a measure $\tilde\mu:\sigma(\mathcal A)\to [0,\infty]$ such that $\tilde \mu(A)=\mu(A)$ for all $A\in \mathcal A$.(Where $\sigma(\mathcal A)$ is the smallest $\sigma$-algebra(or $\sigma$-field) containing $\mathcal A$).
$3.$ Furthermore if $\nu$ is a $\sigma$-finite pre-measure on $\mathcal S$,then the extension to a measure on $\sigma(\mathcal A(\mathcal S))=\sigma(\mathcal S)$ is unique.In fact,$\tilde\mu$ is a $\sigma$-finite measure.
I am not clear about where exactly do we need the $\sigma$-finiteness of the premeasure and what could go wrong in the absence of this extra condition.How would it exactly affect the uniqueness?
Here is a counter-exemple that shows that the assumption of $\sigma$-finiteness is needed to ensure uniqueness.
Let $\Omega=\Bbb Q \cap [0,1)$, $\mathcal{S} = \{\Bbb Q \cap [a,b): a, b \in \Bbb Q\}$ and $\nu:\mathcal S\to [0,\infty]$ the function defined by $\nu(\emptyset)=0$ and $\nu(A)=+\infty$, if $A\neq \emptyset$.
It is easy the check that $\mathcal{S}$ is a semi-algebra and that $\nu$ is a pre-measure. Clearly, $\nu$ is not $\sigma$-finite. It is also easy to check that $\sigma(\mathcal A(\mathcal S))=\sigma(\mathcal S)=\mathcal P(\Omega)$.
Now, let $\#$ be the counting measure and let us define $\tilde\mu_1:\mathcal P(\Omega)\to [0,\infty]$ by $\tilde \mu_1(A) = \#(A)$, and $\tilde\mu_2:\mathcal P(\Omega)\to [0,\infty]$ by $\tilde \mu_2(A) = 2\#(A)$. Clealy, both $\tilde\mu_1$ and $\tilde\mu_2$ are measures that extend $\nu$ to $\mathcal P(\Omega)$. So the extension is not unique.
Remark: Note that both extensions $\tilde\mu_1$ and $\tilde\mu_2$ are themselves $\sigma$-finite. So, in order to have the uniqueness of the extension, the condition that the pre-measure be $\sigma$-finite can not be replaced by the weaker condition that the resulting extension be $\sigma$-finite.
Remark 2: From the point of view of the proof of the uniqueness.
The typical proof goes like that: Suppose there two extensions $\tilde\mu_1$ and $\tilde\mu_2$. Define $$\Sigma = \{ A \in \sigma(\mathcal S) : \tilde\mu_1(A)=\tilde\mu_2(A) \}$$ Clearly, for any $A \in \mathcal S$, $\tilde\mu_1(A)= \nu(A)=\tilde\mu_2(A)$. So $\mathcal S \subseteq \Sigma$. If we can prove that $\Sigma$ is a $\sigma$-algebra, we have $\sigma(\mathcal S) \subset \Sigma$ and it means $\tilde\mu_1=\tilde\mu_2$.
So what remains to be proved is that $\Sigma$ is a $\sigma$-algebra. This is straight forward except for one step: proving that, if $A \in \Sigma$, then $A^c \in \Sigma$. In other words, if $\tilde\mu_1(A)=\tilde\mu_2(A)$, then $\tilde\mu_1(A^c)=\tilde\mu_2(A^c)$. Suppose $\tilde\mu_1(A)=\tilde\mu_2(A)$. We could try
$$ \tilde\mu_1(A^c) = \tilde\mu_1(\Omega)- \tilde\mu_1(A) = \nu(\Omega)- \tilde\mu_1(A) = \nu(\Omega)- \tilde\mu_2(A) = \tilde\mu_2(\Omega)- \tilde\mu_2(A) = \tilde\mu_1(A^c)$$ But the first equality is true for any $A \in \sigma(\mathcal S)$ if and only if $\tilde\mu_1$ is a finite measure. In a similar way, the last equality is true for any $A \in \sigma(\mathcal S)$ if and only if $\tilde\mu_2$ is a finite measure. However, $\tilde\mu_1$ (and $\tilde\mu_2$) is a finite measure if and only if $\nu$ is a finite pre-measure. So, assuming $\nu$ is a finite pre-measure (and that is the critical point), we can complete the proof of uniqueness of the extension.
The next, and last, step is showing that, for the uniqueness of the extension, the case of $\nu$ being a $\sigma$-finite pre-measure can be reduced to the case of $\nu$ being a finite pre-measure. This is done typically by partitioning $\Omega$ in a countable collection subsets where $\nu$ is a finite pre-measure.
(Of course, if $\nu$ is not a $\sigma$-finite pre-measure, then the extensions may not be unique, even if the extensions themselves are $\sigma$-finite , as we have seen in the counter-example above.)