# Is the measure $\nu(A):=\int_A \frac{1}{x}d\lambda$ $\sigma$-finite.

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.

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.

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