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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I could probably write the statement a bit better
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Statement

Suppose that [ilmath]\mu[/ilmath] is either a measure (or a pre-measure) on the [ilmath]\sigma[/ilmath]-ring (or ring), [ilmath]\mathcal{R} [/ilmath] then:

• 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 sigma-subadditivity (or subadditivity) which states that [ilmath]\mu\left(\bigcup_i A_i\right)\le\sum_i\mu(A_i)[/ilmath] (for a pre-measure, 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]
      • By hypothesis, [ilmath]A\in\mathcal{R} [/ilmath], and so is each [ilmath]A_n[/ilmath]. We may invoke a (pre-)measure is sigma-subadditive 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 pre-measures) then:
          • [ilmath]\mu\left(\bigcup^\infty_{n=1}A_n\right)\le\sum^\infty_{n=1}\mu(A_n)[/ilmath]
      • 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 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]
    • 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]
  • 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
• 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.