# Extending Pre-measure into two different measures.

I am trying to find an example of an algebra $$\mathcal{A}$$, and a pre-measure $$\mu_0$$ such that, you can extend $$\mu_0$$ in the $$\sigma-$$algebra generated by $$\mathcal{A}$$ to two different measures.

By Caratheodory's extension theorem, you must have that the trivial extension is not $$\sigma -$$finite.

I was thinking of doing something in $$\mathbb{R}^{\mathbb{R}}$$, and work with integrals (cannot use $$L^p(\mathbb{R})$$ because it is separable), but from that idea, all I have got, are failed attempts.

Any help is appreciated.

• Hmm... maybe $L^\infty (\mathbb{R})$? It's not separable. Commented Aug 9, 2019 at 16:49
• Lebesgue $\sigma$-algebra contains but is different from its Borel $\sigma$-algebra in $\ \mathbb R\.$ Formally, this is an answer (a required example). However, the induced measure in the subalgebra is the same as in the subalgebra. Commented Aug 9, 2019 at 16:56

On $$[0,1)$$, consider the algebra $$\mathcal{A}$$ of all finite unions of half-open intervals, of the form $$\bigcup_{i=1}^n [x_i, x_{i+1})$$. This generates the Borel $$\sigma$$-algebra. Consider the pre-measure $$\mu_0$$ which assigns measure $$+\infty$$ to every non-empty set in $$\mathcal{A}$$. Then one extension of $$\mu_0$$ to the Borel $$\sigma$$-algebra is counting measure $$\mu$$. But $$c \mu$$ is also an extension of $$\mu_0$$ for any $$0 < c \le \infty$$.