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 $\epsilon &gt; 0$ 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| &lt; \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 &gt; 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 $\sum_{i=1}^{\infty}|I_{n_i}| &lt; \epsilon / 2^n$. Therefore, $Y \subset \bigcup_{n,j=1}^{\infty} I_{n_J}$ where $\sum_n \sum_j |I_{n_J}| &lt; \sum_{i=1}^{\infty} \epsilon /2^n = \epsilon$. 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:

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

Like maybe looking at the definition of a series, the limit of the sequence of partial sums. Any help?
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
$$
\sum_n\sum^{\infty}_{j=1}|I_{n_j}|&lt;\sum_n\frac{\epsilon}{2^n}
$$
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
$$\sum_{n=1}^\infty \mu(E_n)=0. $$