$E_1\subset E_2\subset...$ are subsets of $\mathbb{R}$

The limit is defined iff lim sup$E_n$ = lim inf$E_n$.

Here is what I did to prove the double inclusion of the two lim sup and inf:

  1. $\liminf\subset\limsup$ (always true):

Let $x$ be in $\liminf E_n$

$x$ belongs to all $E_n$ except for at most a finite number of them. But $\mathbb{N}\backslash\{i_1,...,i_k\}$ is infinite for a finite $k$ so $x$ belongs to an infinite number of $E_n$ which is an alternative/"intuitive" definition of $\limsup$.

  1. $\limsup\subset\liminf$ (in this particular case):

Let $x$ be in $\limsup E_n$

If $x\notin\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{m\geq n}E_m$ then $\ x\notin\bigcap\limits_{m\geq n}E_m\ \forall n\geq1$ but $\bigcap\limits_{m\geq n}E_m=E_n$ because of how $E_n\subset E_{n+1}\ \forall n$ So $x\notin E_n\ \forall n\in\mathbb{N}\implies x\in\emptyset$

Now I want to prove that mes(lim $E_n$) = lim mes($E_n$) where mes(.) is the measure defined on subsets of $\mathbb{R}$ as $\inf \{\sum\limits_{n=1}^{\infty}(b_n-a_n)\ :\ E\subset\bigcup\limits_{n=1}^{\infty}(a_n,b_n)\}$but I'm not sure how...


1 Answer 1


For pairwise disjunct measurable sets $A_n$ you know that $\rm{mes} (\cup_{m=1}^\infty A_m) = \sum_{m=1}^\infty \rm{mes}(A_m)$ by definition of a measure. An alternative discription of this property is that for any sequence $(E_n)_{n \in \mathbb{N}} \subset \mathcal{A}$ of measurable sets with $E_n \uparrow$, i.e. $E_n \subset E_{n+1}$, one has $\rm{mes}(\cup_{m=1}^\infty E_m) =\lim_{m \rightarrow \infty} \rm{mes}(E_m)$.

The proof can be found in any book on measure theory: Define inductively $A_1 = E_1$ and $A_n = E_n \setminus A_{n-1}$. Then $(A_n)_{n \mathbb{N}}$ are pairwise disjunct, $E_n = \cup_{m=1}^n A_m$, $\cup_{n=1}^\infty E_n = \cup_{m=1}^\infty A_m$, and thus $$ \rm{mes}(\cup_{m=1}^\infty E_m) = \rm{mes}(\cup_{m=1}^\infty A_m) = \sum_{m=1}^\infty \rm{mes}(A_m) = \lim_{n \rightarrow \infty} \sum_{m=1}^n \rm{mes}(A_m) = \lim_{n \rightarrow \infty} \rm{mes}(E_n).$$

  • $\begingroup$ But what guarantees that $\text{mes}$ as defined in the question is indeed a measure? $\endgroup$
    – drhab
    Commented Mar 8, 2018 at 10:47
  • $\begingroup$ The above-mentioned construction of the Borel-measure on $\mathbb{R}$ and the corresponding Borel-$\sigma$-algebra (which for example is generated by all intervals) is known as the "Caratheodory construction". 1.) First, one proves that this defines an outer measure. 2.) Then one defines the set $\mathcal{A}$ of subset of $\Omega$, which satifies the "Caratheodory condition". One can show that this system is a $\sigma$-Algebra. 3.) A regularity condtion (which is satisfied in this the above case) shows that the Borel-$\sigma$-Algebra is a subset of $\mathcal{A}$. Hence $\rm{mes}$ is a measure. $\endgroup$
    – p4sch
    Commented Mar 8, 2018 at 10:58
  • $\begingroup$ Okay. What bothers me is that the OP gives an explicit definition of the measure. That gives the impression that he/she might be still in the middle of proving that it is a measure. If not then it would have been more clear just to say that $\text{mes}$ is a measure, period. $\endgroup$
    – drhab
    Commented Mar 8, 2018 at 11:13

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