Lebesgue measure of union of semi-open interval Given $\mathbf{A} = \bigcup_{n\geq0}[n,n+ \frac{1}{2^n}[$ and the Lebesgue measure $\lambda$, find $\lambda(\mathbf{A})$.
My solution:
\begin{align}
 &\lambda\left(\bigcup_{n\geq0}[n, n+\frac{1}{2^n}[\right) \underset{{additivity\ of \lambda}}{=}\\
&\sum_{n\geq0}\lambda\left([n, n+\frac{1}{2^n}[\right) =\sum_{n\geq0}(n + \frac{1}{2^n} - n) = \sum_{n\geq0}2^{-n} = 2
\end{align}
I'm still confused regarding when to use open, semi-open and closed intervalls, and what kind of semi-open ones, if any. Like what the differences for each type of interval are, when calculating the Lebesgue measure. So I'm not sure, if I shouldn't have changed the original $[n, n+\frac{1}{2^n}[$ interval to some other one like maybe $]n + \epsilon, n+\frac{1}{2^n}-\epsilon]$ (if that's even correct?) and try to work with that.
Sorry if any of my notations are wrong, I'm new to this kind of math. So please correct me if anything I wrote doesn't make sense. 
 A: Remember that with measures, if two measurable sets $A$ and $B$ are disjoint, then $\lambda(A \cup B) = \lambda(A) + \lambda(B)$.
With that in mind, also remember that a singleton set has Lebesgue measure $0$ (can you prove this?).  So for each $x \in \Bbb R$, if $\lambda$ is Lebesgue measure, $\lambda( \{x \}) = 0$.
Using the above two ideas, this means that for any interval $[a,b]$, we have $[a,b] = \{a\} \cup (a,b]$, and since these two sets are disjoint and measurable, we get $\lambda([a,b]) = \lambda(\{a\}) + \lambda((a,b]) = 0 + \lambda((a,b]) = \lambda((a,b])$. This shows that both the intervals $[a,b]$ and $(a,b]$ have the same Lebesgue measure.  The same argument works for $[a,b]$ and $[a,b)$ and also $[a,b]$ and $(a,b)$.  Thus, all of the intervals $[a,b], (a,b], [a,b)$, and $(a,b)$ have the same Lebesgue measure.
So the moral of the story is, as far as Lebesgue measure is concerned, it doesn't matter if you include the endpoints or not, since an endpoint is a singleton set, and singleton sets have Lebesgue measure $0$.
