[proofplan] We first use that $\mathcal{E}$ is a $\sigma$-algebra, so it is closed under countable unions and hence the union is measurable. Then we apply [countable subadditivity](/theorems/1108) of the measure $\mu$ to bound the measure of the union by the infinite sum of the measures of the sets $N_n$. Since every $N_n$ has measure zero, that infinite sum is zero, and non-negativity of measures forces the measure of the union to be exactly zero. [/proofplan]

[step:Show the countable union is measurable] Define the set $N\subset E$ by \begin{align*} N:=\bigcup_{n=1}^{\infty}N_n. \end{align*} For every $n\in\mathbb{N}$, the hypothesis gives $N_n\in\mathcal{E}$. Since $\mathcal{E}$ is a $\sigma$-algebra on $E$, it is closed under countable unions. Therefore \begin{align*} N=\bigcup_{n=1}^{\infty}N_n\in\mathcal{E}. \end{align*} [/step]

[step:Bound the measure of the union by the sum of the measures]Apply countable subadditivity of the measure $\mu$ to the sequence $(N_n)_{n\in\mathbb{N}}$ of measurable sets. This gives \begin{align*} \mu(N)\leq \sum_{n=1}^{\infty}\mu(N_n). \end{align*} For each $n\in\mathbb{N}$, $\mu(N_n)=0$ by hypothesis, so every partial sum satisfies \begin{align*} \sum_{n=1}^{m}\mu(N_n)=0 \end{align*} for every $m\in\mathbb{N}$. The infinite series of non-negative terms is the supremum of its partial sums, hence \begin{align*} \sum_{n=1}^{\infty}\mu(N_n)=0. \end{align*} Therefore \begin{align*} \mu(N)\leq 0. \end{align*}[/step]

[guided]The only estimate needed is countable subadditivity. It applies here because each $N_n$ is measurable, so $(N_n)_{n\in\mathbb{N}}$ is a sequence of sets in the domain $\mathcal{E}$ of the measure $\mu$. Countable subadditivity says that for such a sequence, \begin{align*} \mu\left(\bigcup_{n=1}^{\infty}N_n\right)\leq \sum_{n=1}^{\infty}\mu(N_n). \end{align*} Using the definition \begin{align*} N:=\bigcup_{n=1}^{\infty}N_n, \end{align*} this becomes \begin{align*} \mu(N)\leq \sum_{n=1}^{\infty}\mu(N_n). \end{align*} Now the hypothesis that each $N_n$ is null is used numerically: for every $n\in\mathbb{N}$, \begin{align*} \mu(N_n)=0. \end{align*} Thus for every $m\in\mathbb{N}$, the $m$-th partial sum is \begin{align*} \sum_{n=1}^{m}\mu(N_n)=0. \end{align*} The infinite sum of a non-negative sequence is defined as the supremum of its finite partial sums. Since all partial sums are $0$, the supremum is $0$, and therefore \begin{align*} \sum_{n=1}^{\infty}\mu(N_n)=0. \end{align*} Substituting this into the subadditivity estimate gives \begin{align*} \mu(N)\leq 0. \end{align*}[/guided]

[step:Conclude the union has measure zero] Measures are non-negative, so $\mu(N)\geq 0$. Combining this with the inequality $\mu(N)\leq 0$ gives \begin{align*} \mu(N)=0. \end{align*} We have shown both $N\in\mathcal{E}$ and $\mu(N)=0$. Hence \begin{align*} \bigcup_{n=1}^{\infty}N_n \end{align*} is a null set. This proves the theorem. [/step]