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.


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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.