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Consider the Borel measure, $\nu$, on $[1,\infty)$ given by $$ \nu(A):=\int_A \frac{1}{x}d\lambda $$ where $\lambda$ is the 1-dimensional Lebesgue measure. I want to know if this measure is $\nu$-finite or $\sigma$-finite.

  1. $\nu$-finite. The measure is finite if $\nu([1,\infty))<\infty$. Since $1/x$ is continuous in this domain, the Lebesgue integral coincides with the Riemann integral, and we have $$ \lim_{n\rightarrow\infty}\int_0^n\frac{1}{x}dx=\infty $$ which shows that $\nu$ is not $\nu$-finite.
  2. $\sigma$-finite. Definition from my book Measures, Integrals and Martingales by R. Schilling:
    A measure, $\nu$ is said to be $\sigma$-finite on $(X,\mathscr{A})$ if the $\sigma$-algebra $\mathscr{A}$ contains a sequence $(A_n)_{n\in\mathbb{N}}$ such that $\forall n\in\mathbb{N}:\nu(A_n)<\infty$ and $A_n\uparrow X$, where $$ A_{n} \uparrow X \Longleftrightarrow A_{1} \subset A_{2} \subset A_{3} \subset \ldots \quad \text { and } \quad X=\bigcup_{n \in \mathbb{N}} A_{n} $$ But if $\forall n :A_{n-1}\subset A_n$, then what is the idea of saying that $X=\bigcup_{n \in \mathbb{N}} A_{n}$ instead of $X=\lim_{n\rightarrow \infty} A_n$. And if $\forall n\in\mathbb{N}:\nu(A_n)<\infty$ then I would assume that $X=\bigcup_{n \in \mathbb{N}} A_{n}$ implies that $\nu(X)<\infty$, in which case the definition is the same as the definition in point 1 above.

I would be happy if someone can let me know if my proof in point 1 is correct, and help me with my doubts in point 2.

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1 Answer 1

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I would be happy if someone can let me know if my proof in point 1 is correct, and help me with my doubts in point 2.

Your part 1 looks fine.

But if $\forall n :A_{n-1}\subset A_n$, then what is the idea of saying that $X=\bigcup_{n \in \mathbb{N}} A_{n}$ instead of $X=\lim_{n\rightarrow \infty} A_n$.

When $A_{n-1} \subset A_n$, the limit $\lim_{n \to \infty} A_n$ is defined as $\bigcup_{n \ge 1} A_n$, so they are the same thing.

And if $\forall n\in\mathbb{N}:\nu(A_n)<\infty$ then I would assume that $X=\bigcup_{n \in \mathbb{N}} A_{n}$ implies that $\nu(X)<\infty$, in which case the definition is the same as the definition in point 1 above.

This is not true. For example, the sets $B_n := [-n, n]$ have $\lambda(B_n) = 2n < \infty$ but $\bigcup_{n \ge 1} B_n = \mathbb{R}$ which has $\lambda(\mathbb{R}) = \infty$.

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  • $\begingroup$ Can I pick the sequence $([1,n))_{n\in\mathbb{N}}$? $\endgroup$
    – Hydrogen
    Commented May 9, 2022 at 20:50
  • $\begingroup$ @Hydrogen Yes you may. The definition only mentions existence of such a sequence; if you find one, then you will have verified $\sigma$-finiteness. $\endgroup$
    – angryavian
    Commented May 9, 2022 at 20:51

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