countable sets have outer measure zero

[[concept]]

Topics

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Note

The example from outer measure also means that any countable $A\subset \mathbb{R}$ has $m^{\ast }(A)=0$.

Example

$m^{\ast }(\mathbb{Q})=0$

Proof

If $A$ is countably infinite, then we can write $A=\{a_1,a_2,\dots \}=\{a_n{\}}_{n\in \mathbb{N}}$

Let $ϵ>0$. We will show $m^{\ast }(A)\le ϵ$. For each $n\in N$, let $I_n=\left(a_n-\frac{ϵ}{2^{n+1}},a_n+\frac{ϵ}{2^{n+1}}\right)$

Then

A &\subset \bigcup_{n} I_{n} \\ \implies m^*(A) &\leq \sum_{n=1}^\infty \ell(I_{n}) \\ &= \sum_{i=1}^\infty \frac{\epsilon}{2^n} \\ &= \epsilon \end{align}$$ And since $\epsilon$ was arbitrary, as $\epsilon \to 0$ we get $m^*(A) =0$. $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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