That makes more sense for line segments?

how do we actually define length of curves inside R^2?

Oh, it's refering to paths

@AkivaWeinberger Is my understanding to line correct?

Yeah I guess they mean the concatenation isn't a line

So, $\gamma$ is piecewise linear, and it's broken into as few lines as possible

I see, thanks.

@AkivaWeinberger do you have any idea?

I have many ideas, many of them horrible and few of them relevant

@LeakyNun Defined as a map from an interval to $\Bbb R^2$, or defined as a set of points?

the former

oh alright then just use the parametrization

If that map is differentiable, then it's $\int\|\gamma'(t)\|dt$

If it's not, then you can define it as the supremum of

If $\cal P$ is a partition of $[a,b]$ into $a=t_0,t_1,\dots,t_n=b$, then $\ell(g,\mathcal P)$ is $\sum_{i=1}^n\|g(t_i)-g(t_{i-1})\|$

and then $\ell(g)$ is the supremum of $\ell(g,\mathcal P)$ over all partitions $\cal P$

where $g$ is our curve

@LeakyNun

And then you have to show that that equals $\int_a^b\|g'(t)\|dt$ when $g$ is differentiable

Problem: Suppose $f : E \to (0,\infty)$ is a bounded measurable function such that $\int_E f =0$. Then $m(E)=0$.

I could use a hint on how to solve this problem.

So the important part of that is that $0$ is not in the codomain, I guess. I'll think about this

@user193319 Idea: let $E_n$ be $\{x\in E:f(x)>\frac1n\}$

so it's the subset of $E$ on which $f$ is bigger than $\frac1n$

So then $E=\cup_{n=1}^\infty E_n$. And then we can look at the $\int_{E_n}f$

Oh! Okay. I'll try working with that. Thanks!

One thing I still don't understand is why measure zero set can cause the integral to go to zero even if the function blew up to infinity there

I'm not sure why one needs the bounded hypothesis, in fact. Hmm.

@Ted, quick question for you: why do we teach so much calculus to undergrads?

@AkivaWeinberger why do I smell darboux

Because it's fundamental and useful (and, in the US, at any rate, we teach mostly scientists and engineers (and even business majors), relatively few math majors).

@LeakyNun I don't know, actually

hi, DogAteMy :)

I still forget what Darbu is

I understand why we teach it to non-mathematicians actually

oh just Riemann

Hlello

Because if we look at $\int_{E_n} f d\mu$, won't there be some term k such that the f can overshoot the shrinking of the $E_n$?

But for mathematics students it seems so specific at times

At least in the US, most of the math majors are doing more applied stuff. Only a negligible percentage of math majors go on to grad school in pure math.

I can hardly solve any integral. Maybe someone should have taught calculus to me.

I made my students learn to compute, even in tough theoretical courses, Jakob.

@JakobWerner Ha, that would've helped yeah

This seems to be quite different in Europe, or at least the Netherlands then

But then again I hardly ever needed to solve an integral, so it's fine, I guess

I'll ask a prof at my uni somewhere this week