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Let $\mu$ be a signed measure. Then it has the Jordan decomposition $\mu=\mu^{+}-\mu{-}$. The total variation is defined as $|\mu|=\mu^{+}+\mu{-}$. I am now wondering how to compute an integral $\int f d|\mu|$? I have seen in some other questions here, that it may be defined as $$\int f d|\mu| = \int f d\mu^{+} +\int f d\mu^{-}.$$ I am wondering why this makes sense? Can it also be constructed somehow via simple functions just like the usual Lebesgue integral? Could it also be defined differently?

Are there any books or lecture notes that treat these things in detail?

Edit: So after some further research I found some solutions to Follands book here

https://math24.files.wordpress.com/2013/02/ch3-folland.pdf

There, the author constructs the above integral via simple functions from scratch as follows: Let $\phi = \sum_{i=1}^n a_i \chi_{E_i}$ be a simple function. Then: \begin{align} \int \phi d|\nu|&=\sum_{i=1}^n a_i |\nu|(E_i)=\sum_{i=1}^n a_i (\nu^+(E_i) + \nu^-(E_i))=\sum_{i=1}^n a_i \nu^+(E_i) + \sum_{i=1}^n a_i\nu^-(E_i)\\ &= \int \phi d\nu^+ + \int \phi d\nu^- \end{align}

Now the integral $\int f d|\mu|$ can be computed as follows: \begin{align} \int f d|\mu|=\sup\bigg\{\int \phi d\nu^+ + \int \phi d\nu^- \bigg| \phi \text{ simple and }\phi\leq f\bigg\} \leq \int f d\mu^{+} +\int f d\mu^{-}. \end{align}

So is it really correct that in the line there is an smaller or equal? Or can I get an equal there somehow?

guest1
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  • $\int f d|\mu| = \int f d\mu^{+} +\int f d\mu^{-}$ is a consequence of $|\mu|=\mu^{+}+\mu^{-}$. It is not a definition. – Kavi Rama Murthy Jun 12 '23 at 07:57
  • Hi but how can we see that? This somehow implies that we can do something like $d|\mu|=d\mu^{+}+d\mu^{-}$. But I dont think one can work like this with these infinitesimals... – guest1 Jun 12 '23 at 08:43
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    Prove it for simple functions, then for non-negattive measurable functions, etc. – Kavi Rama Murthy Jun 12 '23 at 08:47
  • Maybe you could have a brief look at https://math24.files.wordpress.com/2013/02/ch3-folland.pdf There they are proving the excercises from Follands book. On p.1 of that document look at exercise 3 a). They do exactly what you have been suggesting, so build it up form simple functions but then they only get $\int f d|\mu| \leq \int f d\mu^- + \int f d\mu^+$ – guest1 Jun 28 '23 at 13:58
  • @geetha290krm could you maybe help me out here? – guest1 Jun 30 '23 at 07:51

1 Answers1

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You have $\nu = |\mu| = \mu^{+} + \mu^{-}$ by definition, which is a sum of mutually singular positive measures. Let $P \perp N$ be a Hahn decomposition for $\mu$, where $P$ and $N$ are disjoint positive and negative sets respectively, so $\mu^{+}(N) = 0$ and $\mu^{-}(P) = 0$.

In other word, $\chi_{P} d |\mu| = d \mu^{+}$ and $\chi_{N} d |\mu| = d \mu^{-}$.

Clearly $f\chi_{P} + f\chi_{N} = f$, and $f^{+} = f^{+} \chi_{P} + f^{+} \chi_{N}$, $f^{-} = f^{-} \chi_{P} + f^{-} \chi_{N}$.

By definition of integral, $\int f d |\mu| = \int f^{+} d |\mu| - \int f^{-} d |\mu| = (\int f^{+} \chi_{P} d|\mu| + f^{+} \chi_{N} d|\mu|) - (\int f^{-} \chi_{P} d|\mu| + f^{-} \chi_{N} d|\mu|) = (\int f^{+} d \mu^{+} + \int f^{+} d \mu^{-}) - (\int f^{-} d \mu^{+} + \int f^{-} d \mu^{-}) = \int f d \mu^{+} + \int f d \mu^{-}$.

$\textbf{Edit}$: I will outline an approach using simple functions that will work.

Assume you have proved this for simple functions.

Write $f = f^{+} - f^{-}$, then $\int f d |\mu| = \int f^{+} d |\mu| - \int f^{-} d |\mu|$.

Let $\phi_n^{+} \uparrow f^{+}$ and $\phi_n^{-} \uparrow f^{-}$ be simple functions monotonically increasing to $f^{+}$ and $f^{-}$ respectively.

Then $ \lim_{n} \int \phi_n^{+} d |\mu| = \int f^{+} d |\mu|$ by the monotone convergence theorem or by definition of integral.

So $\lim_{n} \int \phi_n^{+} d |\mu| = \lim_{n} \int \phi_n^{+} d \mu^{+} + \int \phi_n^{-} d \mu^{-} = \int f^{+} d \mu^{+} + \int f^{+} d \mu^{-}$ again by the monotone convergence theorem or the definition of integral and the hypothesis that the conclusion holds for simple functions.

Do the same thing with $\phi_n^{-}$ and $f^{-}$, and put the results together to get your desired conclusion.

$\textbf{Edit 2}$ : The below response should resolve the matter that troubles you discussed in the comments (namely on splitting the supremum).

Let $ f \geq 0$ and assume you have the desired conclusion for simple functions. Denote simple functions by $\mathcal{E}$.

Then $\int f d |\mu| = \sup \{ \int \phi d |\mu| : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} = \sup \{ \int \phi d \mu^{+} + \int \phi d \mu^{-} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \}$. Clearly $\sup \{ \int \phi d \mu^{+} + \int \phi d \mu^{-} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} \leq \sup \{ \int \phi d \mu^{+} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} + \sup \{ \int \phi d \mu^{-} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} = \int f d\mu^{+} + \int f d\mu^{-}$.

To show the other side, take $0 \leq \phi,\psi \in \mathcal{E} \leq f $ with $\int \phi d \mu^{+} $ and $\int \psi d \mu^{-}$ both within $\frac{\epsilon}{2}$ away from $\int f d \mu^{+}$ and $\int f d \mu^{-}$ respectively.

Then consider $0 \leq \phi \chi_{P} + \psi \chi_{N} \in \mathcal{E} \leq f $ where $P,N$ are as in my post (they form a Hahn decomposition for $\mu$).

Clearly $\int (\phi \chi_{P} + \psi \chi_{N}) d |\mu| = \int \phi d \mu^{+} + \int \psi d \mu^{-} $, and since $(\phi \chi_{P} + \psi \chi_{N})$ is simple, $\int (\phi \chi_{P} + \psi \chi_{N}) d |\mu| \leq \int f d |\mu|$.

But since $\int \phi d \mu^{+} $ and $\int \psi d \mu^{-}$ both within $\frac{\epsilon}{2}$ away from $\int f d \mu^{+}$ and $\int f d \mu^{-}$ respectively, we have that $\int f d\mu^{+} + \int f d\mu^{-} - \epsilon \leq \int f d|\mu|$.

Let $\epsilon \downarrow 0$ to recover $\int f d\mu^{+} + \int f d\mu^{-} \leq \int f d |\mu|$.

This gives the other side.

porridgemathematics
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  • thanks for your answer! Can you maybe comment on my EDIT in my question above? I.e., can you tell me how the construction via simple function works correctly? – guest1 Jul 03 '23 at 13:12
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    @guest1 The construction via simple functions is unnecessary because you can just follow the argument I presented. In any case, the first equality $\int f d |\mu| = \sup ... $ is false for general $f$. It will only work with $f \geq 0$. – porridgemathematics Jul 03 '23 at 13:24
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    On top of that, you should have a $=$ at the end for $f \geq 0$ using the simple function approach. NOT a $\leq$ – porridgemathematics Jul 03 '23 at 13:25
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    I've added an edit which explains how you might go about obtaining your desired conclusion using simple functions. – porridgemathematics Jul 03 '23 at 13:32
  • Hi that with the equal instead of smaller equal is exactly what i dont understand. So in the solution I have found there was a smaller equal and I was confused why there is no equality here. Can you tell me if it would be correct to substitute the smaller equal by an eequal sign? And if so, how can we show that? Can we split the supremum into a sum of supremums or something there to show it more cleary? – guest1 Jul 03 '23 at 13:54
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    You can indeed split the supremum into a sum of supremums, $\textbf{if}$ you assume $f \geq 0$, which renders the $\leq$ a $=$. This is not a problem, because for general $f$, just decompose into positive/negative parts. Its also worth noting that you should add $0 \leq \phi \leq f $ in this case for clarity (although it won't actually make a difference). Overall, because you need to make all of these changes, I wouldn't fixate on the answer provided in that link so much. – porridgemathematics Jul 03 '23 at 13:56
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    @guest1 I've added a final edit which should clarify the matter that troubles you. – porridgemathematics Jul 03 '23 at 14:08
  • great thanks a lot. The only argument that I dont understand is the line that begins with "Clearly, ...", i.e., how you manage to dissect the integral w.r.t. $|\mu|$ into two integrals w.r.t. $\mu^+$ and $\mu^-$? – guest1 Jul 03 '23 at 14:34
  • On a more general note, why do we need the epsilon in that proof? Would it not work without the epsilon too? – guest1 Jul 03 '23 at 14:34
  • Let us continue this discussion in chat. – porridgemathematics Jul 03 '23 at 14:57
  • For others to view these last answers as well I will provide them here below:

    for your first question, this is by assuming the conclusion holds for simple functions. $\int \phi d |\mu|$ can always be split that way because we are assuming that we can do this for $\phi \in \mathcal{E}$

     – guest1 Jul 04 '23 at 07:31
  • for your second question, we need the epsilon in the proof because each of $\int \phi d \mu^{+}$ and $\int \psi d\mu^{-}$ are $\textbf{less}$ than $\int f d \mu^{+}$ and $\int f d \mu^{-}$ respectively (remember we define the latter two integrals in terms of supremums!) so the best we can do is find $\phi, \psi$ simple, such that $\int \phi d \mu^{+} \geq \int f d \mu^{+} - \frac{ \epsilon}{2}$, and the same for $\psi$, and $d \mu^{-}$ I hope this makes sense, try and reread the argument with this in mind – guest1 Jul 04 '23 at 07:31