The (pre)measure of a set is no more than the sum of the (pre)measures of the elements of a covering for that set
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Statement
Suppose that [ilmath]\mu[/ilmath] is either a measure (or a premeasure) on the [ilmath]\sigma[/ilmath]ring (or ring), [ilmath]\mathcal{R} [/ilmath] then^{[1]}:
 for all [ilmath]A\in\mathcal{R} [/ilmath] and for all countably infinite or finite sequences [ilmath](A_i)\subseteq\mathcal{R} [/ilmath] we have:
 [ilmath]A\subseteq\bigcup_i A_i\implies\mu(A)\le\sum_{i}\mu(A_i)[/ilmath]
Note: this is slightly different to sigmasubadditivity (or subadditivity) which states that [ilmath]\mu\left(\bigcup_i A_i\right)\le\sum_i\mu(A_i)[/ilmath] (for a premeasure, we would require [ilmath]\bigcup_i A_i\in\mathcal{R} [/ilmath] which isn't guaranteed for countably infinite sequences)
Proof
Suppose that [ilmath] ({ A_i })_{ i = 1 }^{ n }\subseteq \mathcal{R} [/ilmath] is a finite sequence, in this case we shall consider the countably infinite sequence:
 [ilmath] ({ A_k' })_{ k = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] given by [ilmath]A_k':=A_k[/ilmath] when [ilmath]k\le n[/ilmath] and [ilmath]A_k'=\emptyset[/ilmath] otherwise.
As such we need only prove the statement for infinite sequences (as we implicitly associate each finite sequence with the corresponding infinite sequence by the above construction) (Template:WLOG)
Proof:
 Let [ilmath]A\in\mathcal{R} [/ilmath] be given.
 Let [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] be given such that [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath]
 Define a new sequence: [ilmath] ({ B_n })_{ n = 1 }^{ \infty } [/ilmath] where [ilmath]B_n:=A\cap A_n[/ilmath].
 Notice that [ilmath]\bigcup_{n=1}^\infty B_n=A[/ilmath]^{[Note 1]}
 By hypothesis, [ilmath]A\in\mathcal{R} [/ilmath], and so is each [ilmath]A_n[/ilmath]. We may invoke a (pre)measure is sigmasubadditive which states:
 Given [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] (with [ilmath]\bigcup_{n=1}^\infty A_n\in\mathcal{R}[/ilmath]  which isn't guaranteed to be true for premeasures) then:
 [ilmath]\mu\left(\bigcup^\infty_{n=1}A_n\right)\le\sum^\infty_{n=1}\mu(A_n)[/ilmath]
 Given [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] (with [ilmath]\bigcup_{n=1}^\infty A_n\in\mathcal{R}[/ilmath]  which isn't guaranteed to be true for premeasures) then:
 Applying this to the sequence [ilmath] ({ B_n })_{ n = 1 }^{ \infty } [/ilmath] we see:
 [ilmath]\mu(A)=\mu\left(\bigcup^\infty_{n=1}B_n\right)\le\sum^\infty_{n=1}\mu(B_n)=\sum^\infty_{n=1}\mu(A\cap A_n)[/ilmath]
 Now we have: [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)[/ilmath]
 By monotonicity of (pre)measures^{[Note 2]} we know that:
 [ilmath]\forall A,B\in\mathcal{R}[A\subseteq B\implies\mu(A)\le\mu(B)][/ilmath]
 By the intersection of sets is a subset of each set we know that [ilmath]A\cap A_n\subseteq A_n[/ilmath] (and, less importantly, [ilmath]A\cap A_n\subseteq A[/ilmath])
 Combining these we see that [ilmath]\mu(A\cap A_n)\le\mu(A_n)[/ilmath]
 Thus [ilmath]\sum^\infty_{n=1}\mu(A\cap A_n)\le\sum^\infty_{n=1}\mu(A_n)[/ilmath]^{[Note 3]}
 Combine this with [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)[/ilmath] from before and we see:
 [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A\cap A_n)\le\sum_{n=1}^\infty\mu(A_n)[/ilmath]
 By the transitive property of a partial order, we conclude:
 [ilmath]\mu(A)\le\sum^\infty_{n=1}\mu(A_n)[/ilmath]
 Define a new sequence: [ilmath] ({ B_n })_{ n = 1 }^{ \infty } [/ilmath] where [ilmath]B_n:=A\cap A_n[/ilmath].
 Since [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] was an arbitrary sequence with [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath] we have shown this for all such sequences
 Let [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] be given such that [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath]
 Since [ilmath]A\in\mathcal{R} [/ilmath] was arbitrary, we have shown this for all [ilmath]A\in\mathcal{R} [/ilmath]
This completes the proof.
Notes
 ↑ Should I bother to prove this?
 ↑ Monotonicity is also proved on the premeasure page
 ↑ Adding a proof wouldn't hurt, although it is REALLY obvious, a construction ought to be given though. For any readers following this note because they cannot see the step, note:
 [ilmath]\mu(A\cap A_i)\le\mu(A_i)[/ilmath], then [ilmath]\mu(A\cap A_1)+\mu(A\cap A_2)+\ldots \le \mu(A_1)+\mu(A\cap A_2)+\ldots\le\mu(A_1)+\mu(A_2)+\mu(A\cap A_3)+\ldots[/ilmath] and so forth.
 Remember also that [ilmath]\sum^\infty_{n=1}a_n=\mathop{\text{Lim} }_{n\rightarrow\infty}\left(\sum_{i=1}^n a_i\right)=\mathop{\text{Lim} }_{n\rightarrow\infty}\left(a_1+a_2+\ldots+a_n\right)[/ilmath]
References
