Sattelhofsk

2022-06-21

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We have the outer measure

$$

The inner measure is:

$$

Where ${\mu}_{0}(A)=\sum _{n=1}^{\mathrm{\infty}}{\mu}_{0}({A}_{n})$ and ${\mathcal{A}}_{0}$ is a $\sigma $ algebra.

But why is ${\mu}_{\ast}\le {\mu}^{\ast}$?

I always end up with a contradiction.

My proof would be:

$$

Because:

$$

We get:

$$

$$

Because by definition ${\mu}^{\ast}(A)=inf\sum _{n=1}^{\mathrm{\infty}}{\mu}_{0}({A}_{n})\le \sum _{n=1}^{\mathrm{\infty}}{\mu}_{0}({A}_{n})$ this means, that ${\mu}_{0}(X\mathrm{\setminus}A)-{\mu}^{\ast}(X\mathrm{\setminus}A)\ge 0$

We get:

$$

$$

So where's my mistake?

$$

The inner measure is:

$$

Where ${\mu}_{0}(A)=\sum _{n=1}^{\mathrm{\infty}}{\mu}_{0}({A}_{n})$ and ${\mathcal{A}}_{0}$ is a $\sigma $ algebra.

But why is ${\mu}_{\ast}\le {\mu}^{\ast}$?

I always end up with a contradiction.

My proof would be:

$$

Because:

$$

We get:

$$

$$

Because by definition ${\mu}^{\ast}(A)=inf\sum _{n=1}^{\mathrm{\infty}}{\mu}_{0}({A}_{n})\le \sum _{n=1}^{\mathrm{\infty}}{\mu}_{0}({A}_{n})$ this means, that ${\mu}_{0}(X\mathrm{\setminus}A)-{\mu}^{\ast}(X\mathrm{\setminus}A)\ge 0$

We get:

$$

$$

So where's my mistake?

gaiageoucm5p

Beginner2022-06-22Added 20 answers

${\mu}^{\ast}(A)={\mu}_{0}(A)-{\mu}_{\ast}(\mathrm{\varnothing})$ is incorrect. We have ${\mu}_{0}(A)-{\mu}_{\ast}(\mathrm{\varnothing})={\mu}_{0}(A)-{\mu}_{0}(X)+{\mu}^{\ast}(X)$.