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As we know that a signed measure $\mu$ on $R$ can be decomposed to the positive part $\mu_+$ and negative one $\mu_-$ by the Hahn decomposition theorem.

My question is whether each real-valued Radon measure $\nu$ on $R$ can be decomposed by the positive and negative parts? I think it is true. If the proof is long, Where can I find a reference?

Here is my motivation: the real-valued Radon measure on $R$ seems more general than the signed measure. For example, on the wikipedia, you can find the example $sin(x)d x$ which is a real-valued Radon measure but not a signed measure. By this reason, I would like to work under the real-valued radon measure. However, I am not an expert on measure theory, especially on Radon measure stuff.

Thank you very much!


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2 Answers 2

up vote 4 down vote accepted

I'd take the view that a real-valued Radon measure "really is" a continuous linear functional $\Lambda:C_{00}(X,\mathbb R) \rightarrow \mathbb R$. Here $X$ is a locally compact space, and $C_{00}(X,\mathbb R)$ is the collection of compactly supported continuous real valued functions on $X$. By "continuous", we mean that for each compact $K\subseteq X$, there is a constant $c>0$ such that $$ |\Lambda(f)| \leq c \|f\|_\infty $$ for all continuous $f$ supported in $K$. (For example, with $\Lambda$ given by integrating against $\sin(x) \ dx$, we can take $c$ to be the Lebesgue measure of $K$).

So you could just work abstractly with such linear functionals. (This is going to be a sketch, but if you've done a course on abstract integration, nothing should be too great a surprise). Then you do the usual trick: define $$ \Lambda_+(f) = \sup\{ \Lambda(g) : 0\leq g\leq f \} $$ for all $f\in C_{00}(X,\mathbb R)$ which are (pointwise) positive. The supremum is positive (take $g=0$) and finite because $\Lambda$ is continuous. Notice that $\Lambda_+$ is positive homogeneous, and clearly $\Lambda_+(f_1) + \Lambda_+(f_2) \leq \Lambda_+(f_1+f_2)$. Conversely, if $0 \leq g \leq f_1+f_2$, then let $g_1=\min(g,f_1)$ and $g_2=g-g_1$, so $0\leq g_1\leq f_1$, $0\leq g_2\leq g_2$, and $\Lambda(g) = \Lambda(g_1)+\Lambda(g_2)$. It follows that $\Lambda_+$ is additive.

Then extend $\Lambda_+$ to all of $C_{00}(X,\mathbb R)$ by $\Lambda_+(f) = \Lambda_+(f_+) - \Lambda_+(f_-)$. That $\Lambda_+$ is linear will follow from exactly the same proof which shows that the Lebesgue integral is linear (having first defined it for positive functions) from pretty much any course on integration.

Then define $\Lambda_-(f) = \Lambda_+(f) - \Lambda(f)$, so $$ \Lambda_-(f) = \sup\{ \Lambda(h) : -f\leq h\leq 0 \}\geq 0 $$ (set $h=g-f$). So $\Lambda_+$ and $\Lambda_-$ are positive linear functionals on $C_{00}(X,\mathbb R)$ and are hence given by positive Radon measures.

I don't know a self-contained reference-- as I say, I was exposed to these ideas in courses on integration theory, and basic functional analysis. I like Rudin's book "Real and complex analysis" for the basics on positive Radon measures.

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I think Lang treats this stuff in his "real and functional analysis" book –  Orbicular Jul 15 '11 at 12:31
@Matthew Daws, thank you so much for such detailed answer. It really helps! I still have a doubt. I think $C_{00}(X;R)$ is still a locally convex topology space with semi-norms depending on compact sets? –  Anand Jul 15 '11 at 13:29
@Orbicular, thank you. I will have a look of Lang's book. It is a shame that I have never opened his books. :-) –  Anand Jul 15 '11 at 13:50
@Anand, yes, the notion of continuity is that given by the semi-norms-- this is explained, vaguely, in the Wikipedia article which you link to. –  Matthew Daws Jul 15 '11 at 15:01
@Matthew Daws, yes, I see. Thanks a lot! :-) –  Anand Jul 15 '11 at 17:54

Try Jordan decomposition. A signed measure $\mu$ has positive and negative sets, unique up to measure zero. So do this on each of $[n,n+1)$, then take union, to get positive and negative sets for your signed Radon measure.

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Thanks Professor Gerald Edgar. I happened to come across your book when I looked through measure theory books in our library yesterday. I think it works as you suggested. But we might need a little more care since we need to deal with the restriction of a distribution on a set. This restriction is multiplication of distribution by a indicator function which we couldn't pass to the test function side. But since we only care about zero order distributions, it saves us. Thanks! –  Anand Jul 15 '11 at 13:45

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