[proofplan]
All three properties follow from the countable additivity of $\mu$ and the decomposition of sets into disjoint parts. For monotonicity: $B = A \sqcup (B \setminus A)$, so $\mu(B) = \mu(A) + \mu(B \setminus A) \ge \mu(A)$. Excision follows immediately. For countable subadditivity: "disjointify" the union $\bigcup A_k$ by defining $B_k := A_k \setminus \bigcup_{j<k} A_j$, apply countable additivity to the disjoint union, then bound each $\mu(B_k) \le \mu(A_k)$ by monotonicity.
[/proofplan]
[step:Prove monotonicity]
Let $A, B \in \mathcal{F}$ with $A \subset B$. Write $B = A \sqcup (B \setminus A)$, a disjoint union of two sets in $\mathcal{F}$ (since $B \setminus A = B \cap A^c \in \mathcal{F}$ by closure of $\sigma$-algebras under complements and intersections). By countable additivity of $\mu$ applied to this two-element disjoint union (with all remaining sets equal to $\varnothing$):
\begin{align*}
\mu(B) = \mu(A) + \mu(B \setminus A).
\end{align*}
Since $\mu(B \setminus A) \ge 0$ (measures take values in $[0, \infty]$), $\mu(B) \ge \mu(A)$.
[/step]
[step:Prove excision]
With $A \subset B$ and $\mu(A) < \infty$ as above, $\mu(B) = \mu(A) + \mu(B \setminus A)$. Since $\mu(A) < \infty$, we may subtract it from both sides:
\begin{align*}
\mu(B \setminus A) = \mu(B) - \mu(A).
\end{align*}
The finiteness condition $\mu(A) < \infty$ is essential: without it, the subtraction $\mu(B) - \mu(A)$ could involve $\infty - \infty$, which is undefined.
[guided]
Why require $\mu(A) < \infty$? If $\mu(A) = \infty$, then by monotonicity $\mu(B) \ge \mu(A) = \infty$, so $\mu(B) = \infty$, and $\mu(B) - \mu(A) = \infty - \infty$ is undefined. In this case, $\mu(B \setminus A)$ could be any value in $[0, \infty]$. For example, on $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathcal{L}^1)$, take $A = \{2k : k \in \mathbb{Z}\}$ (measure $0$) and $B = \mathbb{R}$ (measure $\infty$): then $\mu(B) - \mu(A) = \infty - 0 = \infty = \mu(B \setminus A)$, which works. But if we take $A = \mathbb{R}$ and $B = \mathbb{R}$, then $\mu(B) - \mu(A) = \infty - \infty$, while $\mu(B \setminus A) = \mu(\varnothing) = 0$. The excision formula $\mu(B \setminus A) = \mu(B) - \mu(A)$ fails when both sides are infinite.
[/guided]
[/step]
[step:Prove countable subadditivity]
Let $A_1, A_2, \ldots \in \mathcal{F}$ (not necessarily disjoint). Define the "disjointified" sequence
\begin{align*}
B_k := A_k \setminus \bigcup_{j=1}^{k-1} A_j \quad \text{for each } k \in \mathbb{N}.
\end{align*}
Each $B_k \in \mathcal{F}$ (since $\sigma$-algebras are closed under set differences), the $B_k$ are pairwise disjoint (if $k < l$, then $B_l \subset X \setminus A_k \supset X \setminus B_k$, so $B_k \cap B_l = \varnothing$), and
\begin{align*}
\bigcup_{k=1}^\infty B_k = \bigcup_{k=1}^\infty A_k
\end{align*}
(every $x$ in $\bigcup A_k$ belongs to $A_{k_0}$ for some minimal $k_0$, and then $x \in B_{k_0}$). By countable additivity of $\mu$ applied to the disjoint union:
\begin{align*}
\mu\!\left(\bigcup_{k=1}^\infty A_k\right) = \mu\!\left(\bigsqcup_{k=1}^\infty B_k\right) = \sum_{k=1}^\infty \mu(B_k).
\end{align*}
Since $B_k \subset A_k$ for each $k$, monotonicity (proved above) gives $\mu(B_k) \le \mu(A_k)$. Therefore
\begin{align*}
\mu\!\left(\bigcup_{k=1}^\infty A_k\right) = \sum_{k=1}^\infty \mu(B_k) \le \sum_{k=1}^\infty \mu(A_k).
\end{align*}
[guided]
The "disjointification" trick replaces an arbitrary union with a disjoint union having the same total set. The sequence $B_k = A_k \setminus \bigcup_{j < k} A_j$ assigns each point to the first $A_k$ that contains it, ensuring disjointness. Since $B_k \subset A_k$, monotonicity gives $\mu(B_k) \le \mu(A_k)$, and the inequality follows term-by-term.
This technique is fundamental in measure theory: whenever we need to apply countable additivity to a non-disjoint union, we pass to the disjointified sequence and bound with monotonicity.
[/guided]
[/step]
