Proving a weighted sum set function is a measure

Key takeaway: This proof is about verifying the measure axioms directly from the definition, not by appealing to a theorem. The key is to check the value on the empty set and then prove countable additivity for pairwise disjoint sets.

What the definition of a measure requires

To prove that a set function is a measure, you must check two axioms on the given \sigma-algebra:

1. \mu(\varnothing)=0
2. For every pairwise disjoint sequence (A_i), \mu\left(\bigcup_{i=1}^\infty A_i\right)=\sum_{i=1}^\infty \mu(A_i)

In this problem, the function is defined by summing weights of the form 4^i over the index set \Omega A. The important idea is that the measure of a set is determined by which indices belong to it, and the weights are nonnegative.

Step 1: Check the empty set

For the empty set, there are no indices to sum over, so the sum is empty:

$$f(\varnothing)=\sum_{i\in \Omega\varnothing}4^i=0.$$

That gives the first axiom immediately.

Step 2: Use pairwise disjointness

Now let A_1,A_2,\ldots be pairwise disjoint sets. Pairwise disjointness means no element can belong to two different sets at the same time. Therefore, the index set of the union is the disjoint union of the index sets:

$$\Omega\left(\bigcup_{k=1}^\infty A_k\right)=\biguplus_{k=1}^\infty \Omega(A_k).$$

Since the weights 4^i are nonnegative, we may split the sum across disjoint pieces:

$$\begin{aligned} f\left(\bigcup_{k=1}^\infty A_k\right) &=\sum_{i\in \Omega(\bigcup_{k=1}^\infty A_k)}4^i \\ &=\sum_{k=1}^\infty \sum_{i\in \Omega(A_k)}4^i \\ &=\sum_{k=1}^\infty f(A_k). \end{aligned}$$

This is exactly countable additivity.

Why the proof works

The argument does not rely on guessing what the measure "should" be. It uses the defining properties directly. The only facts you need are:

• the empty set contributes no terms,
• disjoint unions separate the sum into non-overlapping parts,
• all terms are nonnegative, so the extended real-valued sum is well behaved.

So the set function satisfies both axioms and is therefore a measure.

Common technical point

If \Omega A is meant to denote the index set describing the elements or atoms of A, the proof should be written in terms of how those indices behave under disjoint unions. If the notation in the source is shorthand, the core logic is still the same: a measure defined as a sum of nonnegative atomic weights is countably additive when the atoms do not overlap.

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Pitfalls the pros know 👇 A common mistake is to treat countable additivity as if it were automatic just because the formula contains a sum. It is not enough to say that both sides are infinite, so they must be equal. You must show that the union decomposes into disjoint parts at the level of the indexing sets, and that the sum over the union equals the sum of the separate sums. Another frequent error is to test only one finite example, such as a set of measure 4, which cannot prove the general axiom. The empty set case also needs a direct argument, not a numerical guess.

What if the problem changes? If the weight function were changed from 4^i to 2^i or c_i for any nonnegative coefficients, the same proof would still work as long as the set function is defined by summing nonnegative weights over disjoint atoms. For example, a variant question might ask: "Let f(A)=\sum_{i\in\Omega A}2^i. Prove f is a measure on \mathcal{P}(\mathbb{R})." The proof is identical: verify f(\varnothing)=0, then show that disjoint unions split the indexing set and the sum. If, however, the weights could be negative, the measure property could fail immediately because measures must take values in [0,\infty].

Tags: countable additivity, pairwise disjoint sets, atomic measure