Is finite measure necessary?

Question

I study http://web.stanford.edu/~eugeniam/math205a/L3.pdf proposition 2.1. Assumes $\nu$ must be finite for absolute continuity iff epsilon delta continuity? Can theorem hold for sigma finite or infinite ? It applies finiteness for the limit theorem but is there example that theorem not true for not finite $\nu$?

Answer 1

Let's try \begin{align} \mu(E) &= \int_E\;dx,\qquad\text{(Lebesgue measure)}\\ \nu(E) &= \int_E x^2\;dx \end{align} Compare two possible definitions for $\nu \ll \mu$:
(a) $\forall E\;\big(\mu(E)=0 \Longrightarrow \nu(E) = 0\big)$
(b) $\forall \epsilon > 0\; \exists \delta > 0\; \forall E\;\big( \mu(E) < \delta \Longrightarrow \nu(E) < \epsilon\big)$

We claim (a) is true but (b) is false.

(b) Take $\epsilon = 1$. We claim there is no $\delta$ which works.
Let $\delta > 0$ be a candidate. Then for any $n > 0$, the set $E = [n,n+\delta/2)$ satisfies $$ \mu(E) = \frac{\delta}{2} < \delta \\ \nu(E) = \int_n^{n+\delta/2} x^2\;dx > \int_n^{n+\delta/2} n^2\;dx = \frac{\delta}{2}\;n^2 $$ There is $n>0$ so that $\frac{\delta}{2}n^2 > 1 = \epsilon$.

Answer 2

How about $X=\mathbb{N}$ and $$\mu(A) = \sum_{k\in A} 2^{-k}$$ $$\nu(A) = \sum_{k\in A} 1$$ ($\nu$ is counting measure and $\mu$ puts weight $2^{-k}$ on the point $k$).

If $\mu(A) = 0$ then we must have $A=\varnothing$ and $\nu(A) = 0$, so it is true that $\nu \ll \mu$.

But the $\varepsilon$-$\delta$-criterion does not hold, e.g. for $\varepsilon=\frac{1}{2}$. (No matter what $\delta>0$ we take, there is some $k$ such that $\mu(\{k\}) = 2^{-k} < \delta$, but $\nu(\{k\}) = 1 > \varepsilon$.)

This example shows that the theorem does not hold for only $\sigma$-finite measures $\nu$.

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The finiteness assumption enters in the proof where the $\sigma$-continuity of $\nu$ from above is used.