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.

  • $\begingroup$ Hmm... maybe $L^\infty (\mathbb{R})$? It's not separable. $\endgroup$
    – Jakobian
    Commented Aug 9, 2019 at 16:49
  • $\begingroup$ 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. $\endgroup$
    – Wlod AA
    Commented Aug 9, 2019 at 16:56

1 Answer 1


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 $0 < c \le \infty$.


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