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?
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$\begingroup$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 $$ \sum_n\sum^{\infty}_{j=1}|I_{n_j}|<\sum_n\frac{\epsilon}{2^n} $$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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 $0$.
$\endgroup$ $\begingroup$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.
$\endgroup$ 6 $\begingroup$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 $$\sum_{n=1}^\infty \mu(E_n)=0. $$
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