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The following theorem has been mentioned (and partially proved) in the book Functions of Bounded Variation and Free Discontinuity Problems by Luigi Ambrosio et. al.

Let $\mu,\nu$ be positive measures on $(X,\mathcal{E})$. Assume that they are equal on a collection of sets $\mathcal{G}$ which is closed under finite intersections. Also assume that there are sets $X_h \in \mathcal{G}$ such that $\displaystyle{X= \bigcup_{h=1}^\infty X_h}$ and that $\mu(X_h) = \nu(X_h) < \infty$ for all $h$. Then $\mu, \nu$ are equal on the $\sigma$-algebra generated by $\mathcal{G}$.

The authors prove the theorem for the case where $\mu,\nu$ are positive finite measures and $\mu(X) = \nu(X)$ and say that the general case follows easily. However, this is not at all straight forward for me. I have tried to prove this but cannot reach a valid proof. Here is my attempt at a proof:

Let $G_h = \{g \cap X_h | g \in G\}$. Then clearly from using the finite case of the theorem, we have that $\mu,\nu$ coincide on every $\sigma (G_h)$. My problem now is to show that this implies that they agree on $\sigma(G)$. All my attempts in this direction have been futile.

I feel that the solution should be relatively easy (as the authors themselves point out). Any help in the proof is greatly appreciated.

Thanks, Phanindra

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  • $\begingroup$ The index $h$ runs over a countable set? $\endgroup$ Jul 11, 2011 at 7:46
  • $\begingroup$ $h$ runs over $\mathbb{N}$ $\endgroup$
    – jpv
    Jul 11, 2011 at 8:08
  • $\begingroup$ changed "$\sigma$-finite" to "general" as $\sigma$-finite tends to imply an increasing family and there is no such restriction on $X_h$ $\endgroup$
    – jpv
    Jul 11, 2011 at 14:37
  • $\begingroup$ jpv: No, your question is still about the $\sigma$-finite case. Whether or not $X_k\subseteq X_{k+1}$ makes no difference. You could assume WLOG that it is increasing if for some reason this helps you, by replacing $X_k$ with $\displaystyle{\cup_{j=1}^k X_j}$. $\endgroup$ Jul 11, 2011 at 17:47

2 Answers 2

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The collection of $A \in \mathcal{E}$ such that $\mu(A \cap X_h) = \nu(A \cap X_h)$ for all $h$ forms a $\sigma$-algebra containing $\sigma(G)$.

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  • $\begingroup$ Let, $H = \{A \in \mathcal{E} | \mu(A \cap X_h) = \nu(A \cap X_h), \forall h\}$. If $A_i \in H \forall i$ then is it true that $\cup_i A_i \in H$? $\endgroup$
    – jpv
    Jul 11, 2011 at 11:26
  • $\begingroup$ Yes, by looking at $(\bigcup_i A_i) \cap X_h = \bigcup_i(A_i \cap X_h)$ for each $h$. $\endgroup$
    – user83827
    Jul 11, 2011 at 16:01
  • $\begingroup$ I see now. Even though $X_h$ are not assumed to be pairwise disjoint, we can create a pairwise disjoint $X_h^\prime$ (as given below by Jonas Meyer) and $\mu(X_h^\prime) = \nu(X_h^\prime)~\forall h$. Then $H$ above is a $\sigma$-field and so $\mu=\nu$ on all of $\sigma(G)$. The only discomfort right now is, where is the finite case of the proof used here? $\endgroup$
    – jpv
    Jul 11, 2011 at 16:26
  • $\begingroup$ I guess perhaps the result isn't used explicitly, but the argument is used in showing the right closure properties when you intersect your sets with each $h$. There's probably some contrived way of rephrasing things to use the finite case literally, but it's probably not as intuitive. $\endgroup$
    – user83827
    Jul 11, 2011 at 16:31
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Note that if $\nu$ is a $\sigma$-finite non negative measure then on a $\sigma$-algebra $G$ then every finite nonnegative measurable function $f$ defines the $\sigma$-finite measure $\mu:=f.\nu$. We can take $f$ to be a characteristic function. This is because, if $f$ is integrable then it defines a set function $\mu(A)= \int_A f d\nu$.

In general, statements involving $\sigma$-finite "situations" can always be reduced to the finite case (see Bogachev section 2.6).

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