My current probability lecture includes the following theorem:
Let $\mathcal{E} \subseteq \mathcal{P}(\Omega)$ be a $\pi$-system which contains a disjoint series $(E_n)_{n \in \mathbb{N}}$ with $\sum_{n=1}^{\infty}E_n = \Omega$. If $\mu_1, \mu_2$ are two measures on $\mathcal{A} := \sigma(\mathcal{E})$ satisfying:
$$ \mu_1(E) = \mu_2(E) \hspace{15px}\forall E \in \mathcal{E} \\ \mu_1(E_n),\,\mu_2(E_n) < \infty\hspace{15px}\forall n \in\mathbb{N} $$
we can already conclude $\mu_1 = \mu_2$ on $\mathcal{A}$.
Now I'm wondering whether the disjointness of the sequence $(E_n)_{n \in \mathbb{N}}$ is really necessary, but cannot come up with a counter example. I'd like to see an example where all the prerequisites except the disjointness of the $E_n$'s are fulfilled, but the final equality doesn't hold.
The disjointness of the sequence is not necessary. This property of the $\mu_i$'s is called $\sigma$-finiteness and it's equivalent if you claim it with or without disjointness (check and compare definition 1. and 2. in the wiki article)
To check this: Assume you have a non-disjoint series $(E_n)_{n\in\Bbb N}$, then consider $$\begin{align*} \tilde{E_0} &= E_0 \\ \tilde{E_n} &= E_n \setminus \bigcup_{k=1}^{n-1} E_k \\ \tilde{\mathcal{E}} &= \mathcal{E} \cup (\tilde{E_n})_{n\in\Bbb N} \end{align*}$$ Now the $\tilde{\mathcal{E}}$ contains a disjoint series with the properties above because the $\tilde{E_n}$'s are disjoint and it holds $$\mathcal{A} = \sigma(\mathcal{E}) = \sigma(\tilde{\mathcal{E}})$$
So we can conclude $\mu_1 = \mu_2$ on $\mathcal{A}$