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Let $\mu$ be a premeasure on the semiring $J$ over $\mathbb{R}$ defined by

$\mu (\emptyset) = 0$ $\mu (I) = \infty \ I \in J, I \not = \emptyset$. Show that there is uncountable many measures when extended to $B(\mathbb{R})$ (borel sets).

We know by theorem that $\mu$ has an extension to a measure $\gamma$ on $\sigma (J)$. But we also know that $\sigma (J) = B (\mathbb{R})$, thus $\gamma$ is a measure on $B( \mathbb{R})$. Aswell I know that $\lambda$ is a measure on the borel sets.

How can I proceed?

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Without any further assumptions on J this in general is false.

Take $J=B(\mathbb{R})$, then J is obviously a semiring but $\mu$ is fixed on $J=B(\mathbb{R})$ so there is just a unique measure that equals $\mu$ on $\sigma(J) = J$: $\mu$ itself.

This is my first post, please forgive me any inconveniences that may come from my inexperience.

So this is how I believe you could proceed:

- Show, that the family of sets $ \mathscr A=\{A\subset\mathbb R \vert \mu(Q)=\mu(Q\cap A)+\mu(Q\setminus A) ,\forall Q\subset \mathbb R\} $ is, in fact a $ \sigma $-Algebra
- Show, that $ \mathscr A_1=\{A\subset\mathbb R \vert \mu(Q)=\mu(Q\cap A)+\mu(Q\setminus A) , Q\subset \mathbb R\} $ (for any $Q\subset \mathbb R$)is a $\sigma$-Algebra
- Argue, that the set $M:=\{Q\vert Q\subset \mathbb R\}$ is uncountable
- Then it follows, that each measure $\mu:\mathscr A_1 \rightarrow [0,\infty]$ is an extention of $\mu$ to the borel sets. Because we showed that there are uncountably many $\mathscr A_1$, the proof is complete.

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