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.

  • 1
    $\begingroup$ It is not clear by what you mean 'when to use ... intervals'. All of the intervals you have above are disjoint, hence $\lambda A = \sum_n { 1\over 2^n} = 2$. The Lebesgue measure is insensitive to single points, hence $\lambda [a,b] = \lambda (a,b)$ and since the half open intervals are sandwiched they have the same measure... $\endgroup$ – copper.hat Apr 23 '16 at 0:38

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 $0$.

  • $\begingroup$ I haven't seen the proof for a singleton. This would be my idea: $\{a\} = [a, a+\epsilon[$ thus $\lambda(\{a\}) = \lambda([a,a+\epsilon[) = a - a +\epsilon = \epsilon$, and for small enough $\epsilon$ this equals 0? $\endgroup$ – Andrei Poehlmann Apr 23 '16 at 0:56
  • $\begingroup$ @reinka Close. You can't say $\{a\} = [a, a + \epsilon)$ because the set on the left hand side contains one element, but on the right hand side the set is an interval, so it has uncountably infinitely many elements. Instead, we can say that for each $\epsilon > 0$, we have $\{a \} \subseteq [a - \frac{\epsilon}{2}, a + \frac{\epsilon}{2}]$. That means $\lambda( \{a \}) \leq \lambda([a - \frac{\epsilon}{2}, a + \frac{\epsilon}{2}])$. But the right hand side equals $a + \frac{\epsilon}{2} - a + \frac{\epsilon}{2} = \epsilon$, so we have $\lambda(\{a\}) \leq \epsilon$. But... $\endgroup$ – layman Apr 23 '16 at 1:03
  • $\begingroup$ @reinka ...Lebesgue measure is nonnegative, so $\lambda(\{a \}) \geq 0$. So we have for every $\epsilon > 0$, $0 \leq \lambda( \{a \}) \leq \epsilon$. Since it's true for every $\epsilon > 0$, this implies $\lambda(\{a\}) = 0$. $\endgroup$ – layman Apr 23 '16 at 1:04
  • $\begingroup$ I see. Very nice explanations of yours. Both regarding your answer and this proof. Really appreciate it! $\endgroup$ – Andrei Poehlmann Apr 23 '16 at 1:08
  • $\begingroup$ @reinka I'm glad you found it useful! $\endgroup$ – layman Apr 23 '16 at 1:29

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