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?
Best Answer
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 $$ \sum_n\sum^{\infty}_{j=1}|I_{n_j}|<\sum_n\frac{\epsilon}{2^n} $$