[asdf_snar]

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.

[ogogmad]

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]

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]

> 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.

[mmmmpancakes]

> 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.

[Tainnor]

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]

Indeed.

[asdf_snar]

Thanks for your reply above.

I don't think the comment was sarcastic or rude. They are pointing out the following inconsistency: you've basically attached "dx" to every integration sign, making the "dx" essentially irrelevant.

Moreover, "dx" does not mean "a small change in x". "dx" is a differential form; it is in particular the "d" operator applied to the function f : x --> x.

As I revisit your comment, I think the point Rota is making about physics notation -- which I _do_ agree with -- is that one should use density functions instead of measures, in general. So, for instance, using the Dirac "density"

\int f(x) \delta(x) dx

instead of

\int f(x) \mu(dx)

where \mu is a point mass at 0. This happens again in the context of stochastic differential equations, where mathematicians shirk away from writing dB_t = xi(t) dt, where xi(t) is "white noise". One can make sense of this in the sense of distributions, and then everything happens in a nice inner product space. Indeed, the physicists are much more competent at actual calculations, and the density representation of things (e.g., in terms of "xi") is very useful for those.

[politician]

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]

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]

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.

[koningrobot]

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?