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Defining multivariable differentiation is just linear algebra. However, defining integration is complicated measure theory. Why are these efforts so different?


marked as duplicate by Parcly Taxel, José Carlos Santos, Cesareo, Ethan Bolker, Vladhagen Oct 23 '18 at 19:43

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    $\begingroup$ @ParclyTaxel It's not a duplicate. The linked question is asking "why is integration harder to perform", this question is asking "why is integration harder to define" $\endgroup$ – 5xum Oct 23 '18 at 9:46
  • $\begingroup$ I suppose the following discussion would be interesting to you: math.stackexchange.com/questions/20578/…. $\endgroup$ – N.B. Oct 23 '18 at 10:41

I don't see these two concepts as so different.

Multivariate differentiation is not "just linear algebra". It involves limits. A strict definition of a differential would be

If $f:\Omega\subseteq\mathbb R^m \to \mathbb R^n$ and $x\in\Omega$, then the differential of $f$ at $x$ is the matrix $D_f(x)\in \mathbb R^{n\times m}$ such that $$f(x+d)=f(x) + D_f(x)\cdot d + o(d)$$ and $$\lim_{d\to 0}\frac{o(d)}{\|d\|} = 0$$

Note that the limits themselves include some "for all $\epsilon>0$, there exists some $\delta>0$" things hidden inside.

On the other hand, integration does not need to involve measure theory. The (arguably) simplest way to define integrals is the Riemann integral which defines the integral as something very similar to a limit. A strict definition of the Riemann integral is

If $f:\Omega\subseteq\mathbb R^m \to \mathbb R$, then the integral of $f$ over $\Omega$ is $L$ if and only if, for every $\epsilon > 0$, there exists some $\delta >0$ such that for every partition $P$ of $\Omega$ into hyperboxes of dimension at most $\epsilon$, the Riemann sum $R_P$ of the function $f$ is at most $\epsilon$-away from $L$, i.e. $$|R_P-L|<\epsilon.$$

I don't see this definition as being much more complicated than the first. It's saying that something very similar to a limit, and actually has no algebra, so it's even simpler.


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