Continuity of a probability measure I am studying Measure Theory and I am stuck in some concepts about continuity of a measure.

Let $(S_{1}, \Sigma,\mu)$ be a measurable space, where $\mu$ is a probability mesure such that $\mu (S_{1}) = 1.$ Let also $S_{\delta} \subset \Sigma, 0\leq \delta \leq 1,$ be a family of sets such that $S_{\delta_{1}} \subseteq S_{\delta_{2}}$ for $0 \leq \delta_{1}\leq \delta_{2} \leq 1.$

Theorem- Continuity from bellow:
 If $E_{j} \subset E_{j+1}$ is an increasing sequence of mesurable sets, then
$$
 \mu\left( \bigcup_{j=1}^{\infty} E_j \right)
  = \lim_{j\rightarrow \infty}\mu(E_{j}).
$$ 

Define $E_{j} = S_{1-1/j}$, for $j \in\{1,2,\ldots\}$.  My question is the following:

If $\mu(S_{\delta}) = c$, with $c &lt; 1$, for $0 \leq \delta &lt; 1,$ how the theorem holds at $\delta =1$ (or $ j\rightarrow \infty$)?

I think I missed something. Thanks in advance!
I see no problem in this.

By the theorem you have $\mu(\bigcup_{j=1}^\infty E_j) = c$ with $\bigcup_{j=1}^\infty E_j = \bigcup_{j=1}^\infty S_{1-1/j} \subseteq S_1$. That $c &lt; 1 = \mu(S_1)$ just tells us that $\bigcup_{0 \leq \delta &lt; 1} S_\delta = \bigcup_{j=1}^\infty S_{1-1/j} \subsetneq S_1$ is a proper subset. (This is what you missed.)

Take for example the unit interval $S_1 = [0,1]$ with the Lebesgue measure $\lambda$ and set $S_\delta = \emptyset$ for every $0 \leq \delta &lt; 1$ (i.e. $c = 0$).
If you define $\mu(S_\delta)=c$ for $c&lt;1$ then $\mu(E_j)=c$ for all $j$, but then $\mu(S_1)=c$ which contradicts with the condition $\mu(S_1)=1$.