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Let $(X, M)$ be a measurable space. If $\nu$ and $\lambda$ are measures on $(X,M)$, define measure $\mu = \nu + \lambda$ to be $\mu(E) = \nu(E) + \lambda(E)$.

Assume $\mu$ and $\nu$ are measures on $(X,M)$ such that $\mu(A) \geq \nu(A)$ for all $A \in M$.

1) If $\nu$ is $\sigma$-finite, show that if $\lambda$ is a measure on $(X,M)$ such that $\mu = \nu + \lambda$ then $\lambda$ is unique. Also give a counterexample to show that this result is false without $\nu $ being $\sigma$-finite

2) Show that there always exists a $\lambda$ such that $\mu = \nu + \lambda$.

I was thinking that to show $\lambda$ is unique, we need to show $\lambda_1(E) = \lambda_2(E)$ for all $E$ if $\nu(E) + \lambda_1(E) = \nu(E) + \lambda_2(E)$. I thought if the sum is finite, then we are done. If the sum is infinite, we somehow want to show $\nu(E)$ is finite. But then I got stuck, I didn't know how to get this from the $\sigma$-finiteness and $\mu \geq \nu$ conditions. I also had a hard time coming up with counter examples. Can someone provide a solution to this problem? Thanks.

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I think the trick to this question is to realize that the intuitive answer of defining $\lambda(E)=\mu(E)-\nu(E)$ would not be well defined as, one might end up with $\infty-\infty$.

Given, $\nu$ is $\sigma-$finite, $\exists F_{i},\ \nu(F_{i})<\infty,\ ,\cup_{i}F_{i}=X$. Now, we can also formulate a partition $\{E_{i}\}_{i\in\mathbb{N}}$ of $X$ by $E_{1}=\emptyset,\ E_{i}=F_{i}\backslash\cup_{j<i}F_{j}$, $\mu(E_{i})\leq\mu(F_{i})<\infty$. We will approach the proof by contradiction. Let there exist measures $\lambda_{1},\lambda_{2}$ which both satisfy $\mu=\nu+\lambda_{i}$, but, they disagree on a set $E,$ i.e, $\lambda_{1}(E)\ne\lambda_{2}(E)$.

Now, $\lambda_{i}(E)=\lambda_{i}(\cup_{i}(E\cap E_{i}))=\sum_{i}\lambda_{i}(E\cap E_{i})$. So, the $\lambda_{i}$'s must disagree on atleast one of the $E\cap E_{j}$. Now, by $\sigma$-finiteness of $\nu$, $\nu(E\cap E_{j})<\infty$. Hence, $\lambda_{1}(E\cap E_{j})\ne\lambda(E\cap E_{j})$, but, $\nu(E\cap E_{j})+\lambda_{1}(E\cap E_{j})=\nu(E\cap E_{j})+\lambda_{1}(E\cap E_{j})$, which is a contradiction.

Now, to see why $\sigma-$finiteness of $\nu$ was crucial for our result, let us define \begin{eqnarray*} \nu(E)=\mu(E) & = & 0\ \ \text{if }E=\emptyset\\ & = & \infty\ \mathrm{otherwise}. \end{eqnarray*}

Now, for arbitrary $x\in X$, and $\forall m>0,$ $\mu(E)=\nu(E)+\lambda(E)$ $ $ and \begin{eqnarray*} \lambda(E) & = & 0\ \ \text{if }x\notin E\\ & = & \infty\ \mathrm{otherwise}. \end{eqnarray*} If, X is singleton, then, we would need : \begin{eqnarray*} \lambda(E) & = & 0\ \ \text{if }x\notin E\\ & = & M\ \mathrm{otherwise\ where\ M\in \mathbb{R}}. \end{eqnarray*}

I am still stuck at part (2).

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  • $\begingroup$ Thanks Math 420 dude. I figured out (ii) though $\endgroup$ – Ron Estrin Sep 26 '14 at 2:06
  • $\begingroup$ Yes, I think I got something too. See you in class tomorrow. I really wish I could work on the problem with others, there are too many assignments, and some of the questions are quite challenging. Did you figure out 4.b.iii ? $\endgroup$ – Juanito Sep 26 '14 at 6:00

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