# Using induction in a measure theory proof

I use $\mu^*$ to denote the outer measure of a subset of $\mathbb{R}$.

Recently on a HW, I had a countable collection of measurable, pairwise disjoint sets {$E_k$}, and I wanted to show $\mu^*(A\cap\bigcup_kE_k)=\sum_k\mu^*(A\cap E_k)$, where $A$ is bounded.

In a previous HW, I proved that $\mu^*(A\cap (E_1\cup E_2))=\mu^*(A\cap E_1)+m^*(A\cap E_2)$.

So I used the latter equation as my base case, and WLOG, since my index set is countable, assumed that the index set for my collection of sets $E_k$ was the set $\mathbb{N}$.

My professor said I am not allowed to use induction here since induction only works for a finite number of objects.

Aren't all the dominoes supposed to fall? That's what the axiom of induction says:

1. $P(i)$ Base Case ($i=2$ in my case)
2. $\forall n. P(n)\Rightarrow P(n+1)$.
3. Then $\forall n.P(n)$

Is induction only allowed for finite sets?

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By the exact same reasoning you can "prove" that $\bigcup_{k\in \mathbb N} \{k\}$ is a finite set (since a union of two finite sets is finite and all the $\{k\}$s are certainly finite). –  Henning Makholm Oct 31 '12 at 23:52
I don't get the title. –  Asaf Karagila Nov 1 '12 at 1:05
Please refrain from using contentless, meme-based titles. –  Potato Nov 1 '12 at 1:40

Right. Though perhaps it's implicit, the axiom of induction should say not just $\forall n$ but $\forall n \in \mathbb{N}$ in its description. The issue here is that infinity (countable or otherwise) is not a natural number. As a side-note, this last observation comes up when young children ask questions about "infinity plus one" and related concepts. Tell them that the "plus" sign should be placed between two numbers, and infinity isn't a number, so it doesn't even make sense to say "infinity plus one." If they're still dissatisfied, put them in front of the wikipage for ordinal arithmetic. –  Benjamin Dickman Nov 1 '12 at 1:23