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I saw another question similar to this one but I'm not satisfied by answers. Here I changed the question to clearify the point I am interested in. I study Measure Theory for Real & Complex Analysis and I wonder why in the following setting

\begin{equation} \mu: \mathcal{M} \rightarrow \mathbb{C} \end{equation} \begin{equation} \mu(E) = \mu_r(E) + i\mu_i(E) \end{equation}

where $\mu$ is a complex measure and $\mu_r$ & $\mu_i$ are signed measure some writers (like Folland if I remember correctly) does not let $\mu$ to attain at most one of $\infty$ or $-\infty$. In the general definition for signed measures we can let signed measures to attain at most one of the plus or negative $\infty$.

For example, is the following setting not meaningful or not sensible?

\begin{equation} \mu: \mathcal{M} \rightarrow \mathbb{C}\cup \{\infty\} \end{equation} \begin{equation} \mu(E) = \mu_r(E) + i\mu_i(E) \end{equation} where $\mu_r$ & $\mu_i$ are signed measures that does not attain $-\infty$, i.e. they are signed measures which attain values in $\mathbb{R} \cup \infty$, and we defined $\mu(E) = \infty$ whenever $\mu_r = \infty$ or $\mu_i = \infty$. Here I rely on following artihmetical definitions in a formal way:

\begin{equation} a + i\infty = \infty = \infty + ai \end{equation}

where $-\infty$ is not considered.

EDIT: Typo in the last equation.

Alileo
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    I don’t think there are any immediate problems with what you wrote down. You can certainly allow a complex measure to take value $\infty$ if you define it carefully. I assume it’s mostly because the reason complex measures are interesting is precisely because they form a vector space over $\mathbb{C}$, and most of the time we only need them because the vector space they form correspond to some other vector space. (For example, the space of complex Radon measures on a compact Hausdorff space is the dual space of the space of continuous functions.) – David Gao Jan 11 '24 at 17:47
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    (Though it does feel a bit weird to say $i \infty = \infty$ whereas the $-i \infty$ is disallowed. Generally speaking, complex infinity does not distinguish between different directions in the complex plane. This might be another reason why complex measures don’t usually allow infinity.) For signed measures, my impression is that they have other uses beyond just considering the space they form, so allowing infinity is more sensible. (And distinguishing between $\infty$ and $-\infty$ is also more sensible to do, given the natural ordering on $\mathbb{R}$.) – David Gao Jan 11 '24 at 17:51
  • @DavidGao The one-point compactification of $\mathbb{C}$ with $\infty$ is what I had in my mind to be honest. The definition I gave in the question tries to justfiy $\infty\cdot (a + ib)$ for any non-zero $a + bi$ which suggest that taking any non-zero complex numbers and scaling it with $\infty$ leads us to $\infty$ in $\mathbb{C}\cup {\infty}$ – Alileo Jan 12 '24 at 15:34
  • @DavidGao By the way I understood the part about the vector space, you can write down an answer instead of a comment. I think I should study topics covering complex measure even more. Later I can think about this definition for one-point compactification of $\mathbb{C}$ and whether or not it is useful for other purposes. – Alileo Jan 12 '24 at 15:58
  • There would be another problem with just using one-point compactification, which is how you deal with infinite summations of finite values that diverge. For example, let assume you have a countable collection of measurable sets all with measure 1 and another countable collection of measurable sets all with measure -1. All these sets are disjoint from each other. The measure of the union of the first collection would be $\infty$, and the measure of the union of the second collection would be $-\infty$. This would be disallowed in the usual definition of signed measures… – David Gao Jan 12 '24 at 20:09
  • But if you are using the one-point compactification, both sums converge to $\infty$, so this seems reasonable. But then you can ask what is the measure of the union of both collections, and there is no value that can satisfy countable additivity. (For example, if you pair up sets from the two collections one-by-one, you get this union is a countable union of measure zero sets, so it should be of measure 0. But it is also the union of two measure $\infty$ sets, so it should be of measure $\infty$ - different ways of pairing the sets can even give you any integer value.) – David Gao Jan 12 '24 at 20:13
  • This is why signed measures explicitly distinguish between $\infty$ and $-\infty$ instead of using the one-point compactification of $\mathbb{R}$, so that this situation can be avoided. In your question the way you did was to restrict the possible ways to achieve infinity within a specific quadrant of the complex plane, but then you do lose the ability to multiply your measures by scalars. Alternatively you can add infinities for all complex directions and enforce that any measure can only achieve infinity in one direction, as I suggested in comments below Tomek’s answer. – David Gao Jan 12 '24 at 20:18

2 Answers2

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Complex measures are a linear space over complex numbers. In order to maintain that with your extension, you would also need to define $a\cdot \infty$ and $a + \infty$, $\infty + \infty$, etc. But there is no good definition of those that maintains $\mathbb{C} + \{\infty\}$ as a linear space. For instance, what would $\infty - \infty$ be?

Edit:

With your definition, you might have a valid extended complex measure, but such that multiplying that measure by a complex constant does not yield a valid extended complex measure. Valid measures aren't closed under multiplications by a constant.

For instance: take the space of natural numbers. Even numbers have measure 1, odd numbers have measure $i$. It's a valid measure by your definition.

But now if you multiply this measure by $1 + i$, it's no longer a valid measure. Now even numbers have measure $1 + i$, odd numbers have measure $-1 + i$, and the sum of real parts for the whole space isn't absolutely convergent.

Tomek Czajka
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    But the same thing is true for signed measure either. – Alileo Jan 11 '24 at 15:41
  • @Alileo I've added an example. Extended signed measures can be multiplied by a constant, but extended complex measures can't always be multiplied by a constant. – Tomek Czajka Jan 11 '24 at 20:58
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    This can seemingly be fixed if we add one $\infty$ for each direction in the complex plane, then allow the measure to take value $\infty$ in exactly one direction. Then multiplication by a finite scalar is perfectly reasonable. Though this feels like a needlessly complicated and unmotivated definition. – David Gao Jan 12 '24 at 00:10
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    @DavidGao Agreed, that does seem to work, but I don't really see why you would really want something like this. Even for real signed measures, often people only allow finite measures. That's how they are defined in Rudin's textbook. – Tomek Czajka Jan 12 '24 at 12:22
  • @TomekCzajka Let's define $z\cdot\infty = \infty$ for non-zero z and $z + \infty = \infty$ for all z and also $\infty + \infty = \infty$. As I explained in the above comment section the one-point compactification is what I had in my mind while I was writing $\mathbb{C}\cup \infty$. – Alileo Jan 12 '24 at 15:37
  • @TomekCzajka I agree. My point is that there seems to be no fundamental difficulties in defining a complex measure that allows infinities. It’s more a conscious choice to not do so, since there seems to be no use at the moment for the notion. – David Gao Jan 12 '24 at 20:06
  • @Alileo Defining it that way you lose the property that a + b + (-1) * b = a, and that's a rather basic property to have. – Tomek Czajka Jan 13 '24 at 22:12
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As suggested, I’m promoting my first two comments to an answer:

I don’t think there are any immediate problems with what you wrote down. You can certainly allow a complex measure to take value $\infty$ if you define it carefully. I assume it’s mostly because the reason complex measures are interesting is precisely because they form a vector space over $\mathbb{C}$, and most of the time we only need them because the vector space they form correspond to some other vector space. (For example, the space of complex Radon measures on a compact Hausdorff space is the dual space of the space of continuous functions.)

Though it does feel a bit weird to say $i\infty = \infty$ whereas the $-i\infty$ is disallowed. Generally speaking, complex infinity does not distinguish between different directions in the complex plane. This might be another reason why complex measures don’t usually allow infinity. For signed measures, my impression is that they have other uses beyond just considering the space they form, so allowing infinity is more sensible. (And distinguishing between $\infty$ and $-\infty$ is also more sensible to do, given the natural ordering on $\mathbb{R}$.)

David Gao
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    Though, now that I think about this more, you can make some complex measures with infinite values into a vector space as well. For example, if you have a locally compact Hausdorff space and you assume some kind of inner regularity and local finiteness on Borel measures, then you can do summation and scalar multiplication on compact sets and then extend to all sets. This would correspond to the dual space of the space of continuous functions with compact support (under a suitable locally convex topology). But quite honestly, one might be better off just sticking to saying it’s the dual space. – David Gao Jan 12 '24 at 20:34