On page 10 of Stefan Grosskinsky's lecture notes, Theorem 1.7 states the following.
There exists an unique Borel measure $\mu$ on $(\mathbb{R}, \mathcal{B})$ such that $\mu((a, b]) = b - a$ for all $a < b$ in $\mathbb{R}$.
The Borel measure claimed to exist is called the Lebesgue measure. In the proof, a set function is defined on the ring of half-open intervals, $$\mathcal{R} := \{\bigcup_{n=1}^{N} (a_n, b_n]: N \in \mathbb{N}, \forall n, a_n < b_n\}.$$ For a set $\bigcup_{n=1}^{N} (a_n, b_n]$ in $\mathcal{R}$ with disjoint intervals, define $$\mu(\bigcup_{n=1}^{N} (a_n, b_n]) := \sum_{n=1}^{N} (b_n - a_n).$$ Clearly this $\mu$ satisfies $\mu((a, b]) = b - a$. The proof then shows that $\mu$ is countably additive. We also know that $\sigma(\mathcal{R}) = \mathcal{B}$, and so by Caratheodory extension theorem, $\mu$ extends to a measure in $\mathcal{B}$.
My question is about the way we prove that $\mu$ is countably additive. In Grosskinsky's lecture notes, this is done by showing that $\mu$ is continuous from above at $\emptyset$. Since we also clearly have $\mu(A) < \infty$ for all $A \in \mathcal{R}$, this implies that $\mu$ is countably additive. However, I've managed to convince myself for the moment that, by defining $\mu$ differently in the first place, we can prove the countable additivity of $\mu$ directly, without resorting to the equivalence between continuity (under certain conditions) and countable additivity. Here it goes.
Define set function $\lambda$ for all unions (finite or countably infinite) of disjoint intervals, $$\lambda(\bigcup_{n \in \mathbb{N}} (a_n, b_n]) := \sum_{n \in \mathbb{N}} (b_n - a_n).$$ Note that $\lambda(\cdot)$ may be $\infty$. This definition is only slightly different from that of $\mu$ above, in that $\lambda$ is defined on possibly (countably) infinite unions of disjoint intervals, whereas $\mu$ is only defined on finite unions of disjoint intervals. We see that $\lambda$ also clearly satisfies $\lambda((a, b]) = b - a$. Moreover, $\lambda$ restricted to $\mathcal{R}$, which we call $\lambda|_{\mathcal{R}}$ for convenience, is the same as $\mu$. The point is that $\lambda$ is countably additive by definition. Consider disjoint sets $(A_n)_{n \in \mathbb{N}}$ in $\mathcal{R}$. The union $\bigcup_{n \in \mathbb{N}} A_n$ is precisely a possibly countably infinite union of disjoint intervals, since each $A_n$ is a finite union of disjoint intervals and the $A_n$ are mutually disjoint. That is to say, $$\bigcup_{n \in \mathbb{N}} A_n = \bigcup_{i \in \mathbb{N}} (c_i, d_i],$$ for some suitable choice of $c_i$ and $d_i$, where the $(c_i, d_i]$ are mutually disjoint. But by definition of $\lambda$, we have $$\lambda(\bigcup_{n \in \mathbb{N}} A_n) = \lambda(\bigcup_{i \in \mathbb{N}}(c_i, d_i]) = \sum_{i \in \mathbb{N}}(d_i - c_i) = \sum_{n \in \mathbb{N}} \lambda(A_n).$$ This is the statement that $\lambda$ is countably additive, as required.
My hunch is that this proof must be wrong, but I can't see what's wrong at the moment. Can you?