# $|A|=\lim_{t\to\infty} |A\cap (-t,t)|,\ A\subset\mathbb{R}, t>0$

I have proved the following statement, but I am not yet sure that everything is correct so I would appreciate comments and corrections, thanks.

$$|A|=\lim_{t\to\infty} |A\cap (-t,t)|,\ A\subset\mathbb{R}, t>0$$

NOTE: $$|\cdot|$$ refers to outer measure, i.e. for $$A\subset\mathbb{R},\ |A|:=\inf\{\sum_{k=1}^{\infty}l(I_k): I_1,I_2,\dots\text{ are open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$$; the length of an open interval $$I\subset\mathbb{R}$$ is defined as

$$l(I):=\begin{cases} b-a & \text{if }I=(a,b),\ a,b\in\mathbb{R}, a

My proof: We study the two possible cases: $$|A|=\infty$$ and $$|A|<\infty$$.

CASE $$|A|=\infty$$| Let $$I_1,I_2,\dots$$ be a sequence of open intervals such that $$A\subset\bigcup_{k=1}^{\infty}I_k$$. Then $$\sum_{k=1}^{\infty}l(I_k)=\infty$$ (since $$|A|\leq\sum_{k=1}^{\infty}l(I_k)$$) so if we take $$M>0$$ arbitrary there must be $$K\geq 1$$ such that $$\sum_{k=1}^{K}l(I_k)>M$$. We note that $$I_k\cap(-t,t)$$ is an open interval and $$A\cap(-t,t)\subset\bigcup_{k=1}^{\infty}(I_k\cap(-t,t))$$ so $$|A\cap(-t,t)|:=\inf\{\sum_{k=1}^{\infty}l(U_k):U_1,U_2,\dots\text{ open intervals such that }A\cap(-t,t)\subset\bigcup_{k=1}^{\infty}U_k\}=\inf\{\sum_{k=1}^{\infty}l(I_k\cap(-t,t)):I_1,I_2,\dots \text{ open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$$ Now, if one the the $$I_k$$ (say, $$I_{k*}$$) is of the form $$(-\infty,a),(a,\infty)$$ ($$a\in\mathbb{R}$$) or $$(-\infty,+\infty)$$ we have that, in the first case $$I_k\cap(-t,t)=(-t,a)\Rightarrow l(I_k\cap(-t,t))=a-(-t)=a+t>M$$ for $$t>\max\{M-a,a\}$$, in the second case $$I_k\cap(-t,t)=(a,t)\Rightarrow l(I_k\cap(-t,t))=t-a>M$$ for $$t>M+a$$ and in the third case $$I_k\cap(-t,t)=(-t,t)\Rightarrow l(I_k\cap(-t,t))=2t>M$$ for $$t>\frac{M}{2}$$ thus in this case it's enough to take $$t>\max\{M-a,M+a,\frac{M}{2}\}$$ to have $$\sum_{k=1}^{\infty}l(I_k\cap(-t,t))\geq l(I_{k*})>M$$. If the $$I_k$$ are all finite ($$I_k=(a_k,b_k)$$) it's enough to take $$t>\max\{|a_1|,|b_1|,|a_2|,|b_2|,\dots, |a_K|,|b_K|\}$$ to have $$\sum_{k=1}^{\infty}l(I_k\cap(-t,t))\geq\sum_{k=1}^{K}l(I_k\cap(-t,t))>M$$. Thus we can conclude that if $$t>\max\{|a_1|,|b_1|,\dots, |a_K|,|b_K|, M-a,M+a,\frac{M}{2}\}$$ we have $$\sum_{k=1}^{\infty}l(I_k\cap(-t,t))>M$$ and taking the $$\inf$$ we get $$|A\cap (-t,t)|>M$$. Since $$M>0$$ was chosen arbitrarily we can conclude that for all $$M>0$$ there exists $$t\geq 0$$ such that $$|A\cap (-t,t)|>M$$ i.e. $$\lim_{t\to\infty} |A\cap (-t,t)|=\infty=|A|$$, as desired.

CASE $$|A|<\infty|$$ since $$|A|=|A\cap (-t,t)|+|A\cap\mathbb{R}-(-t,t)|$$ it is enough to show tha $$\lim_{t\to\infty}|A\cap\mathbb{R}-(-t,t)|=\lim_{t\to\infty}|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|=0$$.

So, let $$I_1,I_2,\dots$$ be a sequence of open intervals ($$I_n =(a_n, b_n)$$ whose union contains $$A$$ (note that there cannot be an infinite one like $$I_{n*}=(-\infty,a)$$, $$(a,\infty)$$ or $$(-\infty,\infty)$$ since in that case $$\sum_{n=1}^{\infty}l(I_n)\geq l(I_{n*})=\infty$$).

For each $$n\geq 1$$ let $$J_n:=I_n\cap (-\infty,-t]$$ and $$L_n:=I_n\cap [t,\infty)$$. Since $$A\cap (-\infty,-t]\cup A\cap [t,+\infty)\subset \bigcup_{n=1}^{\infty} (J_n\cup L_n)$$ we have $$|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|\leq |\bigcup_{n=1}^{\infty}(J_n\cup L_n)|\leq\sum_{n=1}^{\infty}(l(J_n)+l(L_n))=\sum_{n=1}^{\infty}l(J_n)+\sum_{n=1}^{\infty}l(L_n)$$ Now, let $$\varepsilon >0$$: there must be some $$N\geq 1$$ such that $$|\sum_{n=1}^{\infty}l(J_n)-\sum_{n=1}^{N}l(J_n)|=|\sum_{n=N+1}^{\infty}l(J_n)|<\varepsilon/2$$ and similarly there must be $$N'\geq 1$$ such that $$|\sum_{n=1}^{\infty}l(L_n)-\sum_{n=1}^{N'}l(L_n)|=|\sum_{n=N'+1}^{\infty}l(L_n)|<\varepsilon/2$$ so if $$t>\max\{|a_{1}|,\dots,|a_{N}|,|b_{1}|,\dots, |b_{N'}|\}$$ we have that $$\sum_{n=1}^{\infty}l(J_n)+\sum_{n=1}^{\infty}l(L_n)<\varepsilon\Rightarrow |A\cap (-\infty,-t]\cup A\cap [t,+\infty)|<\varepsilon$$. Thus, $$\lim_{t\to\infty}|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|=0$$, as desired.

• Just to clarify: is $|\cdot|$ any outer measure? Mar 23, 2021 at 13:10
• @Idontgetit thank you for your interest in my question; I have edited my question and clarified what I mean by outer measure. Mar 23, 2021 at 13:42

To me it seems that your argument works out if $$A$$ is bounded. To avoid infinite sums, you may then take a finite covering of $$A$$ with open intervals of radius $$\epsilon>0$$ (since the closure of $$A$$ is compact). In other words, you can take the sequence $$(a_n,b_n)$$ such that $$b_n-a_n=\epsilon>0$$ for any $$\epsilon>0$$ and finitely many such intervals suffice to cover $$A$$.
I'm not convinced that your argument works directly for unbounded sets. I think the partial sums are not guaranteed to be bounded by $$\epsilon/2$$.
Here is a similar approach which seems to work generally. The idea is the same, but I choose the intervals $$(a_n,b_n)$$ carefully, to benefit from the result you provided. Since you know: $$|A|=|A\cap (-t,t)|+|A\setminus (-t,t)|$$ It also holds that for $$t'>t$$: $$|A\setminus (-t,t)|=|A\setminus (-t,t)\cap (-t',t')|+|A\setminus (-t,t)\setminus (-t',t')|$$ $$=|A\cap((-t',-t]\cup[t,t'))|+|A\setminus (-t',t')|$$ So, we may write: $$|A|=\sum_{n=0}^\infty|A\cap ((-n-1,-n]\cup[n,n+1))|$$ To see that the sum indeed converges to $$|A|$$, note that if $$|A|=+\infty$$, the sum must diverge and if $$|A|<+\infty$$, then the sum converges monotonely to $$|A|$$. Then, you have for $$t>0$$: $$\sum_{n=0}^{\lfloor t\rfloor}|A\cap ((-n-1,-n]\cup[n,n+1))|\leq |A\cap (-t,t)|$$ $$\leq \sum_{n=0}^{\lceil t\rceil}|A\cap ((-n-1,-n]\cup[n,n+1))|$$ As $$t\rightarrow\infty$$, both the left and right side converge monotonely to $$|A|$$, hence so does $$|A\cap (-t,t)|$$.
• thank you for your answer; I think I might have found another proof for the $|A|=\infty$ case, if you are interested in checking it out. Mar 24, 2021 at 17:49
• Nice Solution! Just a comment to make sure that $$|A|=\sum_{n=0}^\infty|A\cap ((-n-1,-n]\cup[n,n+1))|$$ holds: since the elements of the series are a cover of $A$, $LHS$ is greater than or equal to $RHS$. So, if $|A|=\infty$, the identity trivially holds; and if |A| is finite, since the series is bounded above by $|A|$ (this could be easily concluded from the iterative process forming the series) and its elements are all non-negative, it must converge to a number that is less than or equal to $|A|$. Hence, the reverse inequality holds as well, and the identity is actually valid. Sep 26, 2023 at 7:02