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Inclusion-exclusion for a nonempty finite family of finite-measure sets

Statement

Let m1 be natural and let A0,,Am1 be measurable sets of finite measure. Then

μ(i<mAi)=J{0,,m1}(1)J+1μ(jJAj).

The finite sum on the right uses the following recursive order. For a one-index family it lists {0}. To pass from the order for the nonempty subsets of {0,,m1} to the order for those of {0,,m}, retain the existing list, then append {m}, and then append the sets J{m} for nonempty J{0,,m1} in that existing order. Thus the displayed formula for a family of size m uses only subsets of {0,,m1}. This convention fixes the sum without invoking an unproved permutation rule.

Facts & Assumptions

Given: A nonempty finite list A0,,Am1 of measurable sets, each of finite measure.

[L1]

For measurable A,B, μ(AB)+μ(AB)=μ(A)+μ(B) (The two-set measure identity μ(AB)+μ(AB)=μ(A)+μ(B)).

[L2]

The measure of a finite union is at most the sum of the member measures (Finite and countable subadditivity of measures).

[L3]

Finite sums start with the empty sum and satisfy additivity, splitting, and telescoping laws (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

Natural powers satisfy a0=1 and an+1=ana (Integer powers am).

[L5]

A property true at 0 and inherited by successors holds for every natural number (The principle of mathematical induction).

Proof

technique · induction
1.1

Let P(r) be the displayed inclusion-exclusion formula for a list of r+1 finite-measure measurable sets. It suffices to prove P(r) for every rN.

givenL5
1.2

For r=0, both sides of P(0) are μ(A0), since the only nonempty subset of {0} is {0} and (1)2=1.

baseL3L4
1.3

Fix r and assume P(r) for every list of r+1 such sets.

ih
2.1

Put U=i<r+1Ai. By [L2], μ(U)<+, and [L1] applied to U and Ar+1 gives μ(UAr+1)=μ(U)+μ(Ar+1)μ(UAr+1) in R.

givenstep 1.3L1L2algebra
2.2

The induction hypothesis expands μ(U) over the nonempty subsets of {0,,r} and expands μ(UAr+1)=μ(i<r+1(AiAr+1)) over the same subsets; all intersections remain finite-measure.

givenstep 1.3L2L3
3.1

In step 2.1, the terms from μ(U) are indexed by nonempty subsets not containing r+1, the term μ(Ar+1) is indexed by {r+1}, and the negated terms from step 2.2 are indexed in the stated recursive order by the sets J{r+1} and acquire the sign (1)J+2=(1)J{r+1}+1. Finite-sum splitting therefore gives P(r+1).

step 2.1step 2.2L3L4
4.1

By induction, P(r) holds for every r, hence the stated formula holds for every nonempty finite family; the one-set boundary is step 1.2, and finiteness was used exactly in step 2.1 to permit subtraction.

step 1.1step 1.2step 2.1step 3.1L5discharge-induction

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