Help with a proof that a countable union of measure zero sets is measure zero.

I don't know anything about measure theory, I'm studying real analysis and this showed up in the book I'm reading as a way to characterize integrable functions. The author defined that a subset $X \subset \mathbb{R}$ has measure zero if for each $\epsilon > 0$ we can find infinitely countable open intervals $I_n$ such that $X \subset \bigcup_{n=1}^{\infty}I_n$ and $\sum_{n=1}^{\infty} |I_n| < \epsilon$ where $|I|$ is the size of $I$, as in, if $I = (a,b)$, then $|I| = b - a$.

Now, the author gives the following proof that the countable union of measure-zero sets has measure zero:

"Let $Y =\bigcup_{i=1}^{\infty} X_i$, where each $X_i$ has measure zero. Now, given $\epsilon > 0$ we can, for each $n$, write $X_n \subset \bigcup_{i=1}^{\infty} I_{n_i}$ where each $I_{n_i}$ is an open subset and $\sum_{i=1}^{\infty}|I_{n_i}| < \epsilon / 2^n$. Therefore, $Y \subset \bigcup_{n,j=1}^{\infty} I_{n_J}$ where $\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n = \epsilon$. Therefore, $m(Y) = 0$"

I'm really confused about the ending. It is very intuitive, but it's not rigorous enough for me, I wanna see formally why this holds:

$\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n$

Like maybe looking at the definition of a series, the limit of the sequence of partial sums. Any help?

• If $a_n\leq b_n$ then $\sum_n a_n\leq \sum_n b_n$. That is all is happening there with $a_n=\sum_j|I_{n_j}|$ (<-- bad notation by the way) and $b_n=\epsilon/2^n$.
– Tom
Feb 20 '15 at 0:35
• @raffa Sorry about that, I deleted my answer. Feb 20 '15 at 0:55

Given two sequences $x_n$ and $y_n$, we have the relation $$x_n\leq y_n\Rightarrow \sum_n x_n\leq \sum_n y_n.$$ Therefore, if we set $x_n=\sum^{\infty}_{j=1}|I_{n_j}|$ and $y_n=\frac{\epsilon }{2^n}$, applying the above gives $$\sum_n\sum^{\infty}_{j=1}|I_{n_j}|<\sum_n\frac{\epsilon}{2^n}$$

We're exploiting Zeno's Paradox. Since each set has measure $0$, we can cover it by intervals whose total length is less than any positive real number. Since the union is countable, we can enumerate our sets of measure $0$ as $\{I_1, I_2, I_3, \ldots, \}$. Let $\mu(S) = (b-a)$ for $S=(a,b)$.

Let $\epsilon > 0$. Let $I_j$ be covered by an open covering so that $$\mu(A_j) < \frac{\epsilon}{2^j}$$ $$I_j \subset A_j$$ Then, since $\mu$ is countably sub-additive, $$\mu\left(\bigcup_{j=1}^{\infty} I_j \right) \leq \sum_{j=1}^{\infty} \mu(I_j) \leq \sum_{j=1}^{\infty} \mu(A_j) = \sum_{j=1}^{\infty} \frac{\epsilon}{2^j} = \epsilon.$$

Thus, our set has measure 0.

This is nice because we're able to do a countable number of ''things'' with only an epsilon of room.

That's just how measures work. We start by defining that the measure of an open interval $(a,b)$ is $b-a$. Then we can attempt to define the (outer) measure of an arbitrary set $A$ as the infimum of all $\sum_{i\in J}\mu(I_i)$ where the $I_j$ are open intervals and $A\subseteq \bigcup_{i\in J}I_i$. A few observations:

• If more than countably many of the $\mu(I_i)$ are nonzero, then certainly $\sum_{i\in J}\mu(I_i)=\infty$. Therefore and as any set $A\subseteq \mathbb R$ allows a countable cover, we nay restrict to the case that $J$ is countable.
• With the extended definition, we still have $\mu((a,b))=b-a$. One shows by induction that $\sum_{i\in J}\mu(I_i)\ge b-a$ if $A\subseteq \bigcup_{i\in J}I_i$ with finite $J$, and then extends this to the case of countable $J$ (and larger $J$ need not be considered)

Now given countably many $X_k$, $k\in\mathbb N$, with $\mu(X_k)=0$, and given $\epsilon>0$, by definition as infimum we find covers $X_k\subseteq \bigcup_{j\in J_k}I_{k,j}$ with $\sum_{j\in J_k}\mu(I_{k,j})<2^{-k}\epsilon$. By the above we may assume that $J_k=\mathbb N$. Then $$X:=\bigcup_{k\in\mathbb N}X_k\subseteq\bigcup_{k\in\mathbb N}\bigcup_{j\in\mathbb N}I_{k,j}=\bigcup_{(k,j)\in\mathbb N\times\mathbb N}I_{k,j}$$ with $$\sum_{(k,j)\in\mathbb N\times\mathbb N}\mu(I_{k,j})=\sum_{k\in\mathbb N}\sum_{j\in\mathbb N}\mu(I_{k,j})<\sum_{k\in\mathbb N}2^{-k}\epsilon=\epsilon$$ so that $\mu(X)<\epsilon$. As $\epsilon$ was an arbitrary positive number, the infimum over all $\sum_{i\in J}\mu(I_i)$ for covers $X\subseteq \bigcup_{i\in J}\mu(I_j)$ is certainly $0$.

Equalities and inequalities with series of series are preserved under exchanging order of summation indices and also taking inner limits before outer limits, vs. summing all the terms in any order you want, so long as the series are either all positive or all negative terms. Thus you don't have to worry about the double summation and how it affects the inequality.

• Yeah, I kind of get that, but I'm getting confused about the specifics of it all. I'm trying to come up with a rigorous proof, with a specific list of theorems that I'm using and everything. Feb 20 '15 at 0:52
• @raffa In that list none of the facts he mentions is necessary to include.
– Tom
Feb 20 '15 at 0:56
• @Tom I was thinking about what you wrote, and I got what you said, but I don't think that's quite enough. Now I'm worried about the fact that the series we are supposed to show that is $< \epsilon$ is actually equal to $\sum_n \sum_j |I_{n_J}|$ Feb 20 '15 at 0:59
• @raffa You should probably pinpoint the exact first step that you don't understand in the proof. The sum $\sum_n\sum_j|I_{n_j}|$ is the one that turns out to be $<\epsilon$, yes. That is the last line in the proof and also the definition of $Y$ having measure zero.
– Tom
Feb 20 '15 at 1:40
• I've pinpointed it! I agree with you, I understand the inequality and why it holds. Now, my problem is before that. It's this: I think it needs to be shown that the infinite sum of all the sizes of the intervals converges and is equal to $\sum_n \sum_j |I_{n_J}|$ Feb 20 '15 at 1:44

In general, if $(X,\mathcal M,\mu)$ is a measure space and $\{E_n\}$ is a sequence of sets in $\mathcal M$, let $$F_n=E_n\setminus\bigcup_{j=1}^{n-1}E_j.$$ Then for any $n$, $$\bigcup_{j=1}^n F_n=\bigcup_{j=1}^n E_n$$ and $F_n\cap F_m=\varnothing$ for $m\ne n$. So by countable additivity, $$\mu\left(\bigcup_{n=1}^\infty E_n\right) = \mu\left(\bigcup_{n=1}^\infty F_n\right) = \sum_{n=1}^\infty \mu(F_n).$$

Now, as $E_n$ is the disjoint union of $E_n\cap F_n$ and $E_n\setminus F_n$, we have $$\mu(E_n) = \mu(E_n\cap F_n) +\mu (E_n\setminus F_n) = \mu(F_n)+\mu(E_n\setminus F_n)\geqslant \mu(F_n).$$ It follows that $$\sum_{n=1}^\infty \mu(F_n)\leqslant \sum_{n=1}^\infty \mu(E_n)$$ and since $\mu(E_n)=0$ for all $n$, that $$\sum_{n=1}^\infty \mu(E_n)=0.$$

• That uses like a million things I don't know... I'm trying to focus on series and elementary analysis. Feb 20 '15 at 1:05
• This is actually a much simpler approach to the problem. I'll edit the answer to add some detail though. Feb 20 '15 at 1:07
• I didn't read that you didn't know anything about measure theory. In that case, you should probably study some general measure theory before trying to tackle Lebesgue measure/integration. Feb 20 '15 at 1:14
• @raffa You don't need to learn any of that to to understand your proof. Actually, this answer shows that even those that know some measure theory don't know how to give a simple proof of the theorem. This proof is unnecessarily involved.
– Tom
Feb 20 '15 at 1:35
• This holds in general measure spaces. I don't see how you think it is "unnecessarily involved" - these are all basic facts which would be proved in an introductory graduate-level real analysis course, which is the level at which the question was asked... Feb 20 '15 at 1:48