How to prove a set function is a measure and find its total mass

Expert intro: This kind of problem checks two things at once: whether you can verify the axioms of a measure, and whether you can read the definition carefully enough to compute the total value without overthinking it. The key is that the function only “sees” natural numbers inside a set.

Detailed walkthrough

Let

$\mu(A)=\sum_{i\in A\cap\mathbb N}\frac1{2^i},\qquad A\subseteq\mathbb R.$

We verify the measure axioms on $(\mathbb R,\mathcal P(\mathbb R))$.

1) Nonnegativity

Each term $\frac1{2^i}$ is nonnegative, so every sum defining $\mu(A)$ is nonnegative. Hence

$\mu(A)\ge 0 \quad \text{for all }A\subseteq\mathbb R.$

2) Null empty set

Since

$\varnothing\cap\mathbb N=\varnothing,$

we get

$\mu(\varnothing)=\sum_{i\in\varnothing}\frac1{2^i}=0.$

3) Countable additivity

Take pairwise disjoint sets $(A_n)_{n=1}^\infty\subseteq\mathcal P(\mathbb R)$. For each natural number $i$, the indicator of membership satisfies

$\mathbf 1_{\cup_n A_n}(i)=\sum_{n=1}^\infty \mathbf 1_{A_n}(i),$

because the sets are disjoint, so each integer belongs to at most one $A_n$. Therefore

=\sum_{i\in (\cup_n A_n)\cap\mathbb N}\frac1{2^i} =\sum_{i\in\mathbb N}\Bigl(\sum_{n=1}^\infty \mathbf 1_{A_n}(i)\Bigr)\frac1{2^i}.$$ Because all terms are nonnegative, we may rearrange the sums: $$=\sum_{n=1}^\infty \sum_{i\in A_n\cap\mathbb N}\frac1{2^i} =\sum_{n=1}^\infty \mu(A_n).$$ So $\mu$ is countably additive. Since $\mu$ is nonnegative, null on the empty set, and countably additive, it is a measure. ### 4) Compute $\mu(\mathbb R)$ Now $$\mathbb R\cap\mathbb N=\mathbb N,$$ so $$\mu(\mathbb R)=\sum_{i=1}^\infty \frac1{2^i}.$$ This is a geometric series with first term $\frac12$ and ratio $\frac12$: $$\mu(\mathbb R)=\frac{\frac12}{1-\frac12}=1.$$ So the answer is: $$\boxed{\mu \text{ is a measure and } \mu(\mathbb R)=1.}$$ ### 💡 Pitfall guide A common slip is to forget that the sum only runs over $A\cap\mathbb N$, not over all of $A$. Real numbers that are not natural numbers contribute nothing. Another easy mistake is writing $\sum_{i\in\mathbb R}$, which is not what the definition says and is not a meaningful series here. ### 🔄 Real-world variant If the coefficient were changed to $\frac{1}{3^i}$, the same proof would still work, and the total mass would become $$\sum_{i=1}^\infty \frac1{3^i}=\frac{1/3}{1-1/3}=\frac12.$$ If the indexing set were $\mathbb Z_{>0}$ instead of $\mathbb N$, nothing essential changes as long as the same positive integers are used in the sum. The measure is still purely atomic, with atoms at the natural numbers. ### 🔍 Related terms countably additive measure, atomic measure, geometric series