Theorem $:$ Let $\mathcal C$ be a semi-algebra of subsets of $X$ and $\mathcal A (\mathcal C), \mathcal S (\mathcal C)$ respectively denote the algebra and the $\sigma$-algebra of subsets of $X$ generated by $\mathcal C.$ Let $\mu_1,\mu_2 : \mathcal S (\mathcal C) \longrightarrow [0,+\infty]$ be two finitely additive set functions on $\mathcal S (\mathcal C)$ such that $$\mu_1 (E) = \mu_2 (E),\ \ \text {for all}\ E \in \mathcal C.$$ Then for all $E \in \mathcal {A} (\mathcal C)$ $$\mu_1 (E) = \mu_2 (E).$$
The above theorem follows from the facts that $\mu_1$ and $\mu_2$ are finitely additive and $$\mathcal A (\mathcal C) = \left \{E \subseteq X\ \bigg |\ E = \bigsqcup_{i=1}^{n} C_i,\ C_i \in \mathcal C, n \in \Bbb N \right \}.$$
Now we need to prove the extension theorem for measure with the help of the above theorem.
Extension theorem for measure
Let $\mathcal C$ be a semi-algebra of subsets of $X.$ Let $\mathcal S (\mathcal C)$ denote the $\sigma$-algebra of subsets of $X$ generated by $\mathcal C.$ Let $\mu_1,\mu_2 : \mathcal S (\mathcal C) \longrightarrow [0,+\infty]$ be two $\sigma$-finite measures on $\mathcal S (\mathcal C)$ such that $$\mu_1 (E) = \mu_2 (E),\ \ \text {for all}\ E \in \mathcal C.$$ Then for all $E \in \mathcal S (\mathcal C)$ $$\mu_1 (E) = \mu_2 (E).$$
To prove this theorem WLOG we may assume that $\mathcal C$ is a semi-algebra because by previous theorem we already know that $\mu_1 (E) = \mu_2 (E),$ for all $E \in \mathcal A (\mathcal C)$ and we also know that $\mathcal S (\mathcal A (\mathcal C)) = \mathcal S (\mathcal C)$ and hence the assertion follows.
In my book the second simplification what has been made is that "WLOG we may also assume that both $\mu_1$ and $\mu_2$ are totally finite measures" which is not quite clear to me. What I have tried is as follows $:$
As far as I understood the problem it turns out to me that in order for second assertion to hold the only thing we need to show is that the theorem holds for $\sigma$-finite measures as long as it holds for totally finite measures. So suppose that the theorem holds for any pair of totally finite measures and we need only to show that it holds for $\sigma$-finite measures $\mu_1$ and $\mu_2$ given in the theorem.
Since both $\mu_1$ and $\mu_2$ are $\sigma$-finite measures $\exists$ partitions $X_1,X_2, \cdots$ and $Y_1,Y_2, \cdots$ of $X$ such that $X = \bigsqcup\limits_{i=1}^{\infty} X_i = \bigsqcup\limits_{j=1}^{\infty} Y_j$ with $\mu_1 (X_i) < +\infty$ and $\mu_2 (Y_j) < + \infty$ and $X_i,Y_j \in \mathcal S (\mathcal C),$ for all $i,j \in \Bbb N.$ Let $E_{ij} = X_i \cap Y_j,$ for all $i,j \in \Bbb N.$ Then $X = \bigsqcup\limits_{i=1}^{\infty} \bigsqcup\limits_{j=1}^{\infty} E_{ij}$ and $\mu_1 (E_{ij}) < + \infty$ and $\mu_2 (E_{ij}) < + \infty,$ for all $i,j \in \Bbb N.$ Let $\mu_1^{(ij)}$ and $\mu_2^{(ij)}$ respectively denote the restriction of $\mu_1$ and $\mu_2$ to $\mathcal S (\mathcal C) \cap E_{ij} = \mathcal S (\mathcal C \cap E_{ij}),$ for all $i,j \in \Bbb N.$ Then both $\mu_1^{(ij)}$ and $\mu_2^{(ij)}$ are totally finite measures on $\mathcal S (\mathcal C \cap E_{ij}),$ for all $i,j \in \Bbb N.$ Now let $A \in \mathcal S (\mathcal C).$ We need to show that $\mu_1 (A) = \mu_2 (A).$ Now $A = \bigsqcup\limits_{i=1}^{\infty} \bigsqcup\limits_{j=1}^{\infty} (A \cap E_{ij})$ and hence \begin{align*} \mu_1 (A) & = \sum\limits_{i=1}^{\infty} \sum\limits_{j=1}^{\infty} \mu_1 (A \cap E_{ij}) \\ & = \sum\limits_{i=1}^{\infty} \sum\limits_{j=1}^{\infty} \mu_1^{(ij)} (A \cap E_{ij}). \end{align*} So if we can show that $\mu_1^{(ij)} (A \cap E_{ij}) = \mu_2^{(ij)} (A \cap E_{ij}),$ for all $i,j \in \Bbb N$ then we are through. Now $A \cap E_{ij} \in \mathcal S (\mathcal C \cap E_{ij}),$ for all $i,j \in \Bbb N.$ So if we can somehow show that $\mu_1^{(ij)}$ and $\mu_2^{(ij)}$ agree on $\mathcal C \cap E_{ij}$ for all $i,j \in \Bbb N$ then we are through by our assumption since both $\mu_1^{(ij)}$ and $\mu_2^{(ij)}$ are totally finite measures. What we only know is that $\mu_1$ and $\mu_2$ agree on $\mathcal C.$ I find difficulty to prove this part. Can anybody please help me in this regard?
Thank you very much for your valuable time for reading.
You have trouble proving your statement because it is not true as stated.
As an example, consider the semi-algebra of all non-empty half-open intervals $$ C = \{\emptyset, \Bbb{R}\} \cup \{[a,b) : a < b\}. $$ Note that $C$ generates the Borel $\sigma$-algebra $B$. Now, define $\mu, \nu : B \to [0,\infty]$ by $$ \mu(M) = \# (M \cap \Bbb{Q}) \quad \text{and} \quad \nu(M) = \# (M \cap[\Bbb{Q} + \sqrt{2}]). $$ It is easy to see that $\mu(\emptyset) = 0=\nu(\emptyset)$ and $\mu([a,b)) = \infty = \nu([a,b))$ for all $a<b$, so that $\mu,\nu$ agree on $C$. Furthermore, $\mu,\nu$ are $\sigma$-finite, since $\Bbb{Q}$ and $\sqrt{2} + \Bbb{Q}$ are countable. Nevertheless, $\mu \neq \nu$.
To get a correct statement, you can assume that $\mu,\nu$ are $\sigma$-finite with respect to $C$, meaning that $X = \bigcup_n M_n$ with $M_n \in C$ and $\mu(M_n) < \infty$. I will leave the proof to you.