Theorem $:$ Let $\mathcal A$ be an algebra of subsets of a non-empty set $X.$ Let $\mu : \mathcal A \longrightarrow [0, + \infty]$ be a finitely additive set function with $\mu (\varnothing) = 0.$ Then $\mu$ is countably additive if and only if the following condition holds $:$
For any $A \in \mathcal A$ $$\mu (A) = \lim\limits_{n \to \infty} \mu (A_n)$$ whenever there exists a sequence $\{A_n \}_{n \geq 1}$ in $\mathcal A$ with $A_n \subseteq A_{n+1},$ $\forall$ $n \geq 1$ such that $A = \bigcup\limits_{n=1}^{\infty}A_n.$
Proof $:$ Suppose $\mu$ is countably additive. Now take any $A \in \mathcal A$ and a sequence $\{A_n \}_{n \geq 1}$ in $\mathcal A$ with $A_n \subseteq A_{n+1},$ $\forall$ $n \geq 1$ such that $A = \bigcup\limits_{n=1}^{\infty} A_n.$ Define $B_1 := A_1$ and $B_n : = A_n \setminus \bigcup\limits_{i=1}^{n-1} A_i,$ $\forall$ $n \gt 1.$ Then the sequence $\{B_n \}_{n \geq 1}$ has the property that $B_n \cap B_m = \varnothing$ for $n \neq m$ and $\bigcup\limits_{n = 1}^{k} B_n = \bigcup\limits_{n=1}^{k} A_n,$ for every $k \geq 1.$ Hence taking limit as $n \to \infty$ it follows that $A = \bigcup\limits_{n=1}^{\infty} A_n = \bigcup\limits_{n=1}^{\infty} B_n.$ Since $B_n$'s are pairwise disjoint it follows from countable additivity and finite additivity of $\mu$ that \begin{align*} \mu (A) & = \mu \left (\bigcup\limits_{n=1}^{\infty} B_n \right ) \\ & = \sum\limits_{n=1}^{\infty} \mu (B_n) \\ & = \lim\limits_{k \to \infty} \sum\limits_{n=1}^{k} \mu (B_n) \\ & = \lim\limits_{k \to \infty} \mu \left (\bigcup\limits_{n=1}^{k} B_n \right ) \\ & = \lim\limits_{k \to \infty} \mu \left ( \bigcup\limits_{n=1}^{k} A_n \right ) \\ & = \lim\limits_{k \to \infty} \mu (A_k). \end{align*} The last equality follows from the fact that $A_n \subseteq A_{n+1},$ $\forall$ $n \geq 1.$ This proves one part of the theorem.
To prove the converse let us assume that the given condition holds. Now let us take a sequence $\{A_n \}_{n \geq 1}$ in $\mathcal A$ such that $A_n \cap A_m = \varnothing,$ for $n \neq m.$ Let $A = \bigcup\limits_{n=1}^{\infty} A_n \in \mathcal A.$ Let us define $B_n : = \bigcup\limits_{k=1}^{n} B_k,$ $\forall$ $n \geq 1.$ Then the sequence $\{B_n \}_{n \geq 1}$ has the property that $B_n \subseteq B_{n+1},$ $\forall$ $n \geq 1$ and $A = \bigcup\limits_{n=1}^{\infty} A_n = \bigcup\limits_{n=1}^{\infty} B_n.$ So by the given condition and finite additivity of $\mu$ it follows that \begin{align*} \mu(A) & = \lim\limits_{n \to \infty} \mu (B_n) \\ & = \lim\limits_{n \to \infty} \mu \left ( \bigcup\limits_{k=1}^{n} A_k \right ) \\ & = \lim\limits_{n \to \infty} \sum\limits_{k=1}^{n} \mu (A_k) \\ & = \sum\limits_{k=1}^{\infty} \mu (A_k). \end{align*}
This proves the reverse part of the theorem and this completes the proof.
In the entire proof I have understood everything except the fact that where have used the fact that $\mu (\varnothing) = 0.$ Would anybody please help me in this regard?
Thank you very much for your valuable time for reading.
As far as I can tell, the proof does not use the fact thet $\mu(\emptyset)=0$ (or that $X\ne\emptyset$), and I'm not too scared of the possibility of missing something, because it is not needed for the statement to be true. In fact, given an algebra of sets $\mathcal U$ (but really any family of sets containing $\emptyset$) and an additive function $\mu:\mathcal U\to[0,\infty]$, then $$\mu(\emptyset)=\mu(\emptyset\cup\emptyset)=2\mu(\emptyset) $$
which is true if and only if $\mu(\emptyset)=0\lor\mu(\emptyset)=\infty$. If the second, then for all $A\in\mathcal U$ $$\mu(A)=\mu(\emptyset\cup A)=\infty+\mu(A)=\infty$$
It is clear that the constant $\infty$ function is both $\sigma$-additive and continuous from below (which is the name by which that property the lemma mentions sometimes goes).
However, it is worth mentioning that most of the time, in the literature, a measure is required to satisfy $\mu(\emptyset)=0$ one way or another.