=== Necessary Condition === To show is that a measure $\mu$ has the properties $(1)$, $(2)$, $(3)$, $(3')$ and $(3'')$. Property $(1)$ is also part of the definition of measure, and hence is immediate. Property $(2)$ is precisely the statement of Measure is Finitely Additive Function. Next, let $\sequence {E_n}_{n \mathop \in \N} \uparrow E$ in $\Sigma$ be an increasing sequence. Define $F_1 = E_1$, and, for $n \in \N$: :$F_{n + 1} = E_{n + 1} \setminus E_n$ Then as $\Sigma$ is a $\sigma$-algebra: :$\forall n \in \N: F_n \in \Sigma$ Also, the $F_n$ are pairwise disjoint as $\sequence {E_n}_{n \mathop \in \N}$ is an increasing sequence. By construction, have for all $k \in \N$ that: :$\ds E_k = \bigcup_{n \mathop = 1}^k F_n$ and so: :$\ds E = \bigcup_{n \mathop \in \N} F_n$ Hence, as $\mu$ is a measure, compute: {{begin-eqn}} {{eqn | l = \map \mu E       | r = \map \mu {\bigcup_{n \mathop \in \N} F_n}       | c = by the above reasoning }} {{eqn | r = \sum_{n \mathop = 1}^\infty \map \mu {F_n}       | c = $\mu$ is a Measure }} {{eqn | r = \lim_{k \mathop \to \infty} \sum_{n \mathop = 1}^k \map \mu {F_n}       | c = {{Defof|Series}} }} {{eqn | r = \lim_{k \mathop \to \infty} \map \mu {\bigcup_{n \mathop = 1}^k F_n}       | c = Measure is Finitely Additive Function, Finite Union of Sets in Additive Function }} {{eqn | r = \lim_{k \mathop \to \infty} \map \mu {E_k}       | c = by the above reasoning }} {{end-eqn}} This establishes property $(3)$ for measures. For $(3'')$, note that it is a special case of $(3')$. For property $(3')$, let $\sequence {E_n}_{n \mathop \in \N} \downarrow E$ be a decreasing sequence in $\Sigma$. Suppose that $\map \mu {E_1} < +\infty$. By Measure is Monotone, this implies: :$\forall n \in \N: \map \mu {E_n} < +\infty$ and also: :$\map \mu E < +\infty$ Now define: :$\forall n \in \N: F_n := E_1 \setminus E_n$ Then: :$F_n \uparrow E_1 \setminus E$ {{explain|prove this claim}} Hence, property $(3)$ can be applied as follows: {{begin-eqn}} {{eqn | l = \map \mu {E_1} - \map \mu E       | r = \map \mu {E_1 \setminus E}       | c = Measure of Set Difference with Subset }} {{eqn | r = \lim_{n \mathop \to \infty} \map \mu {E_1 \setminus E_n}       | c = by property $(3)$ }} {{eqn | r = \lim_{n \mathop \to \infty} \paren {\map \mu {E_1} - \map \mu {E_n} }       | c = Measure of Set Difference with Subset }} {{eqn | r = \map \mu {E_1} - \lim_{n \mathop \to \infty} \map \mu {E_n} }} {{end-eqn}} Here, all expressions involving subtraction are well-defined as $\mu$ takes finite values. It follows that: :$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$ as required. {{qed|lemma}} === Sufficient Condition === The mapping $\mu$ is already satisfying axiom $(1)$ for a measure by the imposition on its codomain. Also, axiom $(3')$ is identical to assumption $(1)$. It remains to check axiom $(2)$. So let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$. Define, for $n \in \N$: :$F_n = \ds \bigcup_{k \mathop = 1}^n E_k$ Then: :$\forall n \in \N: F_n \subseteq F_{n + 1}$ Also, by Additive Function is Strongly Additive: :$\ds \forall n \in \N: \map \mu {F_n} = \map \mu {\bigcup_{k \mathop = 1}^n E_k} = \sum_{k \mathop = 1}^n \map \mu {E_k}$ Hence, using condition $(3)$ on the $F_n$, obtain: {{begin-eqn}} {{eqn | l = \map \mu {\bigcup_{n \mathop \in \N} E_n}       | r = \lim_{n \mathop \to \infty} \map \mu {F_n}       | c = Condition $(3)$ }} {{eqn | r = \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \map \mu {E_k}       | c = by the reasoning above }} {{eqn | r = \sum_{k \mathop = 1}^\infty \map \mu {E_k}       | c = {{Defof|Series}} }} {{end-eqn}} This establishes that $\mu$ also satisfies axiom $(2)$ for a measure, and so it is a measure. Now to show that $(3')$ and $(3'')$ can validly replace $(3)$. As $(3')$ clearly implies $(3'')$ (which is a special case of the former), it will suffice to show that $(3'')$ implies $(3)$. {{finish|this really isn't hard (use set difference and finiteness on decreasing sequence $G_n :{{=}} \bigcup_{k \mathop \in \N} E_k \setminus F_n \downarrow \O$) but I can't formulate it nicely; the case where {{LHS}}, {{RHS}} are infinite for axiom $(2)$ has to be covered. Feel free to finish it.}} {{qed}}