# Why is the Lebesgue Outer Measure of $\mathbb{R}$ equal to $+\infty?$

If $A\subseteq \mathbb{R}$, then the Lebesgue outer measure of $A$, denoted $m^*(A)$, is defined to be $$m^*(A)=\operatorname{inf}\left\{\sum_{k=1}^{+\infty}\ell(I_k)\right\}$$ where the infimum is taken over all countable collections $\{I_k\}$ of open intervals with the property that $A\subseteq \bigcup_{k=1}^{+\infty}I_k.$ Using the definition above, how do we show that $m^*(\mathbb{R})=+\infty$? One thing comes to my mind is to show that for any open covering $\{I_k\}$ of $\mathbb{R}$, $\sum_{k=1}^{+\infty}\ell(I_k)=+\infty$, of which I got a difficulty of showing it. Is there a simple way of explaining that $\mathbb{R}$ has infinite outer measure?

Note the Lebesgue outer measure is monotone ($A \subseteq B$ implies $m^* (A) \leq m^* (B)$), and I assume you know that $m^* ( [-a,a] ) = 2a$ for all $a > 0$.

Put these together and you get that $m^* ( \mathbb{R} ) \geq 2a$ for all $a > 0$, which implies that $m^* ( \mathbb{R} )$ cannot be a finite value.

• Thanks. Monotonicity and Comparison Properties are the main tricks!!! Thank you very much sir... May 1 '13 at 6:33
• @juniven. You're welcome. (You could prove it straight form the definition, but it would be much more work. It's better to work smarter rather than harder!) May 1 '13 at 6:46
• you're absolutely right. I like your last statement. May 1 '13 at 6:52

Try showing $m^* [-M,M] \ge 2M$. Since $[-M,M] \subset \mathbb{R}$, this will give the desired result.

Since $[-M,M]$ is compact, any open cover has a finite subcover.

• Thanks. All of you gave precise and compact answers. May 1 '13 at 6:47

Outer measure is monotone so the outer measure of $\mathbb{R}$ must be larger than or equal to the outer measure of all its subsets. Condier the interval $[-n,n]\subset\mathbb{R}$ which has outer measure $2n$. So

$$2n=m^*([-n,n])\le m^*(\mathbb{R})$$

for all natural numbers $n$. So it has infinite outer measure.

• If I can only accept all answers, then I will also accept yours. Thanks for the answer. May 1 '13 at 6:49