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I think I am one of these mathematicians that doesn't understand the logic. How can I write μ(x)dx instead of μ(dx) without risking the confusion that dx is Leb

I think I am one of these mathematicians that doesn't understand the logic. How can I write μ(x)dx instead of μ(dx) without risking the confusion that dx is Lebesgue measure? You may have explained this in your other reply, but I don't quite follow. P.S. Beautiful integral signs.

politician 4y ago
Please excuse my imposition, as I am a humble programmer who spends his days adding and subtracting 1, and not a mathematician. Yet, this discussion of the confusion and potential confusion of misinterpreting notation strikes me as something that has long (well, in the sense of programming) been solved in my area with type systems and syntax highlighting. Do mathematicians not have these tools?
- Sharlin 4y ago
  There is no syntax highlighting on the blackboard. Math notation is not designed. It has haphazardly evolved over centuries. It is not rigorous even though math itself is (attempts to be) rigorous, it is a language as imperfect as its users. But it does its job well enough.
- ABeeSea 4y ago
  They do! It’s called abstract algebra and it’s very similar to type theory in a lot of ways. But to get to the rigorous mathematician definition of manipulating dx and dy, it requires a large amount of the machinery from abstract algebra that’s hard to quickly absorb or explain.
ogogmad 4y ago
I'm suggesting maybe writing the Lebesgue measure as 1 so the Lebesgue integral of a function f becomes ⌠ | f(x)⋅1 dx ⌡ The logic is that the Lebesgue measure is a density which is everywhere equal to 1. Given a measurable space over \mathbb R^n, I think there is only one such measure. Another example is that δ(x) in ⌠ | f(x)⋅δ(x) dx ⌡ represents the Dirac measure. For producing this ASCII art, I use Sympy. I write for instance pprint(Integral(f(x) * delta(x)))
- mmmmpancakes 4y ago
  So what does dx mean in this setting then? If the answer is nothing, then let me suggest simplifying your expression to the following: ⌠ | f(x)⋅μ(x) ⌡ Now it occurs to me that the only problem with this new notation is that you risk confusing which term is the density (especially if there are multiple greek letters floating around). To clarify this potential confusion I have a solution! Add some notation to indicate which is the density: ⌠ | f(x)⋅dμ(x) ⌡ Wait...
  - ogogmad 4y ago
    > So what does dx mean in this setting then? It means a small change in x. It means the same thing in dy/dx. So you know who GCR was, but this is the first time you're hearing of this... > Wait... So your comment's sarcastic. You're a rude anonymous maths fan.
    - Tainnor 4y ago
      I think you're the rude person. GP wasn't being sarcastic, they were showing you why they think that taking your notational convention to its logical conclusion, you end up with the notation we have today.
      - mmmmpancakes 4y ago
        Indeed.
    - mmmmpancakes 4y ago
      > It means a small change in x. So here is the issue. "A small change in x" is a concept that is relevant and meaningful for Riemann integration: it represents that the integral is defined as a limit of Riemann sums as the change in x becomes infinitely small. But this is not relevant for Lebesgue integration! We are not taking a limit of Riemann sums and there is not a limit of change in x getting arbitrarily small. What matters is the measure (and the definition of the integral in terms of simple functions is something entirely different). So you see using dx this way in the context of Lebesgue integration seems like a potential source of confusion, not simplification. This is why i ask what dx means. Either it represents the standard Lebesgue measure on R (this is valid and a special case of the notation d\mu(x) since \mu is the identity function), or it is nonsense that potentially confuses concepts from other integration theories. This is why mathematicians don't like your idea. They encounter many students who are confused about this distinction and don't find that it's a useful simplification.
koningrobot 4y ago
Did you mean to write dμ(x) instead of μ(dx)? As a non-measure-literate person, I can understand dμ(x) as the d of density μ(x) evaluated at x. But μ(dx) has μ evaluated at the infinitesimal dx which is very different. Is μ(dx) the correct notation?

politician 4y ago

Please excuse my imposition, as I am a humble programmer who spends his days adding and subtracting 1, and not a mathematician. Yet, this discussion of the confusion and potential confusion of misinterpreting notation strikes me as something that has long (well, in the sense of programming) been solved in my area with type systems and syntax highlighting. Do mathematicians not have these tools?

Sharlin 4y ago
There is no syntax highlighting on the blackboard. Math notation is not designed. It has haphazardly evolved over centuries. It is not rigorous even though math itself is (attempts to be) rigorous, it is a language as imperfect as its users. But it does its job well enough.
ABeeSea 4y ago
They do! It’s called abstract algebra and it’s very similar to type theory in a lot of ways. But to get to the rigorous mathematician definition of manipulating dx and dy, it requires a large amount of the machinery from abstract algebra that’s hard to quickly absorb or explain.

ogogmad 4y ago

I'm suggesting maybe writing the Lebesgue measure as 1 so the Lebesgue integral of a function f becomes ⌠ | f(x)⋅1 dx ⌡ The logic is that the Lebesgue measure is a density which is everywhere equal to 1. Given a measurable space over \mathbb R^n, I think there is only one such measure. Another example is that δ(x) in ⌠ | f(x)⋅δ(x) dx ⌡ represents the Dirac measure. For producing this ASCII art, I use Sympy. I write for instance pprint(Integral(f(x) * delta(x)))

mmmmpancakes 4y ago
So what does dx mean in this setting then? If the answer is nothing, then let me suggest simplifying your expression to the following: ⌠ | f(x)⋅μ(x) ⌡ Now it occurs to me that the only problem with this new notation is that you risk confusing which term is the density (especially if there are multiple greek letters floating around). To clarify this potential confusion I have a solution! Add some notation to indicate which is the density: ⌠ | f(x)⋅dμ(x) ⌡ Wait...
- ogogmad 4y ago
  > So what does dx mean in this setting then? It means a small change in x. It means the same thing in dy/dx. So you know who GCR was, but this is the first time you're hearing of this... > Wait... So your comment's sarcastic. You're a rude anonymous maths fan.
  - Tainnor 4y ago
    I think you're the rude person. GP wasn't being sarcastic, they were showing you why they think that taking your notational convention to its logical conclusion, you end up with the notation we have today.
    - mmmmpancakes 4y ago
      Indeed.
  - mmmmpancakes 4y ago
    > It means a small change in x. So here is the issue. "A small change in x" is a concept that is relevant and meaningful for Riemann integration: it represents that the integral is defined as a limit of Riemann sums as the change in x becomes infinitely small. But this is not relevant for Lebesgue integration! We are not taking a limit of Riemann sums and there is not a limit of change in x getting arbitrarily small. What matters is the measure (and the definition of the integral in terms of simple functions is something entirely different). So you see using dx this way in the context of Lebesgue integration seems like a potential source of confusion, not simplification. This is why i ask what dx means. Either it represents the standard Lebesgue measure on R (this is valid and a special case of the notation d\mu(x) since \mu is the identity function), or it is nonsense that potentially confuses concepts from other integration theories. This is why mathematicians don't like your idea. They encounter many students who are confused about this distinction and don't find that it's a useful simplification.

koningrobot 4y ago

Did you mean to write dμ(x) instead of μ(dx)? As a non-measure-literate person, I can understand dμ(x) as the d of density μ(x) evaluated at x. But μ(dx) has μ evaluated at the infinitesimal dx which is very different. Is μ(dx) the correct notation?