Does "non-differentiability" imply "non-integrability"

Continuity is a sufficient condition for integrability. It is not necessary. Go back to calc 2

>It is not necessary.
You are right but wrong. I mistated my theorem.

A function is integrable if and only if it is bounded and has only a countable number of discontinuities.

If the function is "continuous nowhere" then the set of discontinuities is uncountable, therefore it is not integrable.

>A function is integrable if and only if it is bounded and has only a countable number of discontinuities.
This isn't correct either

Consider

[eqn] f(x) = \begin{cases} 1
& \text{ if } x \in \mathbb{Q} \\
0 & \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} [/eqn]

That's clearly an integratable function but it's nowhere continuous since the rationals are dense in the real numbers.

>that's clearly an integrable function

It's a step function. You do know that integrals are constructed by first defining them for all measurable step functions, right?

I haven't seem a calculus shitstorm this intense since pic related

The function is integrable at c if and only if the limit as x approaches c of f(x) is equal to f(c).

This doesn't work because the limit as x approaches c from either side does not exist, regardless of what c is. Therefore it is not integrable.

yeah, that's what i thought, so why is differentiation and integrations considered inverses?

[math]f(x) = | \, x \, |[/math]
[math] \nexists \frac{\text{d} f(x= 0)}{\text{d} x}[/math]
but with [math]a0[/math]
[math]\int_a^b f(x) \text{d}x = \int_a^0-x \text{d} x +\int_0^bx \text{d} x [/math]
How bout dah

FUCKIN UNDERGADN NORMIES GET OFF MY BOARD REEEEEEEEEEEE

Learn what a Lebesgue integral is before spouting shit, please.

>quantum level

They aren't direct inverses. They're only related by the FTC. That's it.

Aren's step functions finite linear combinations of delta functions? (I think in English the function that's 1 in a set and 0 outside is a delta function, but I'm not sure at all).

In any case, this would be a limit of step functions, so you're still right.

If they are only related by the FTC then, the FTC doesn't sound so fundamental anymore.

Well, just think that for any function [math]g[/math] on [math]\Omega[/math] that satisfies

[math]g = \int _\Omega f(x) dx[/math]

integration and derivation serve as (pseudo)inverses. So for the world of [math]C^k \, \, \, k>0[/math], the FTC is a statement about (pseudo)inverse operations.

today I learned Veeky Forums doesnt even know how to integrate a simple function.

>Lebesgue integral

Please only talk about things that actually exist, like the riemann integral.

hgahaha xd very funnyny hahah he is denying some well known mathematical object form 1000 years agoo ahahahha
also infintite and reals doesnt reall ajajajaja xdxdxdxddd yopu are so funny wilderberger is my her xddd reals noooo jaja rationals are twehere is it infinity desont real

you're presumably talking about riemann integrability. "Integrable" in this context typically refers to the Lebesgue integral.

fucking brainlet fucking kill yourself fuck you

no

A function is riemann integrable if the set of discontinuities has lebegue measure 0 ( it needs to be bounded obviously)

>Lebesgue integral
That's just a Riemann integral but with horizontal slices instead of vertical right? I still can't understand how that can integrate that clusterfuck of a function.

>"Integrable" in this context typically refers to the Lebesgue integral.
There is no context at all. Integrable can mean with regard to any damn type you want if you don't specify and it's natural to use the simplest one in these cases, because if OP doesn't specify he probably doesn't know or care about other integrals.

> Integrable can mean with regard to any damn type you want if you don't specify

But then questions like "Is the indicator function of Q integrable?" are ambiguous, not to mention the answer to OP's question depends on the meaning of "integrable". There are multiple ways OP's question could be interpreted, and I think it's worthwhile to point out that the word "integrable" has a different meaning to someone with an advanced knowledge of math than it does to someone who just finished Calc I.

>clearly an integratable function
>clearly
OK von Nerdmann show us how to integrate that function over some interval.

fyi, a function is integrable if its discontinuities have Lebesgue measure zero. Since [math]\mathbb{Q}[/math] has measure 0 (because it's countable, I'm not going to explain to you what a null set is, but it's basically a set in [math]\mathbb{R}^n[/math] with nonzero codimension) we can say this function is 0 "almost everywhere," so if we were going to integrate this function over the compact set [math][a,b][/math], we can just forget its discontinuities like this:

[math]\int _a ^b f(x)dx = \int _a ^b 0dx = 0[/math]

But he is right. We should consider the Riemann integral here. But the answer is that the function doesn't need to be differentiable to be integrable. I don't know why OP asked for this becouse he has the anwer in his picture

>Quantum level

Can't you just find the area under the graph?

>Does "non-differentiability" imply "non-integrability"
>post an example of a function that's nowhere differentiable but obviously integrable

>but obviously integrable
uhhh

All continuous functions are integrable.

How, if it's not well-defined?
By "well-defined," I mean that you can keep "zooming in."

Are you gonna try to argue that the rect() function ain't integrable, brainlet?

That's not what well-defined means, and it is well defined.
And what do you mean you can "zoom in". You can zoom into most graphs. integrable or not
And it's because it's a continuous function, and FTC says continuous functions are integrable.

do some analysis dude

I know that's not what well-defined means; it has an equation consisting of elementary functions, of course it's well-defined.

I mean that if you were to do a Riemann sum, you might find that the length of one of the bars is 1. But then you "zoom in" and you see it's 1.05. Keep zooming in, wait, it's 1.025. Now, I know that the function evaluated at a certain point has a certain value independent of the spooky graph, but once you start integrating--moving along the curve--you can't reliably measure the lengths and sum them.

I'm sure I'm mistaken, I just want you to be able to understand my mistaken thought process to explain where I went wrong in my reasoning

You are not just mistaken but also retarded my friend.

Why are you in an analysis thread if you haven't even taken calc I? Do you even know the definitions of derivative and integral?

The function value doesn't change from 'zooming'

Look at the curve, it's obvious that any section has a finite area under it. So if you take the limit of the Riemann sums it will converge to something, even if the edges go a bit crazy.

I swear every time I come crossboarding here to Veeky Forums it's all threads like this where most of the posters are literally braindead. Embarrassing frankly

fpbp

For all you people stating the obvious examples like the rect function, I don't think OP means functions that are not differentiable at a point or region. I think he's talking about functions like the Weierstrass function which aren't differentiable anywhere...
Taking the integral is essentially the same thing as taking the area under the curve of the function. However, this isn't easy to do under a fractal curve.

Fortunately, integrals are actually rather easy. You can do term by term integration over this sum:
math.stackexchange.com/questions/829911/what-does-the-antiderivative-of-a-continuous-but-nowhere-differentiable-function

So the simple answer is "no, 'non-differentiability' does not imply 'non-integrability' because you can easily find the antiderivative of a non-differentiable function like the Weierstrass function".

Not sure if b8, but [math]f[/math] is discontinuous everywhere, so it's not Riemann integrable.

actually Lebesgue integral is considered canonical. Riemann integration is still taught because it's easier to construct, but any time you write the integral sign "without specifying the integral", Lebesgue integral is the implicit choice. it's superior and there is no reason to not prefer it.

Q has measure 0, therefore R\Q is also measurable and has the complementary volume, namely b-a, if we restrict to an interval. now you can tear the integral into two integrals over measurable sets, the function becoming constant on each of them. therefore it's clearly integrable.

THIS IS THE RIGHT ANSWER

No. There are extremely many integrable functions which are not differentiable.

Jesus christ, Lebesgue integrals are the standard integrals. Don't spout shit if you're still an undergrad. Look up measure theory and Lebesgue integration please.

Because they are, the same way the successor is the inverse of the predecessor.
That doesn't that mean 0 has a predecessor because it has a successor.

While you are right for this particular function, this argument :
> it's nowhere continuous since the rationals are dense in the real numbers.
is wrong.

Considere the same function except when x is rational then f(x)=1/q
q defined by x=p/q the most simplified fraction.

Then this new function is discontinuous on all rationals and CONTINUOUS on all irrationals.
So despite being discontinuous on a dense set, it is continuous almost everywhere.

I'd like to code this and represent it graphically (just for curiosity). Obviously not to infinity.

But what is a? Just some sort of tuning parameter I can set to whatever?

That's a neat fact.

I'm guessing a and b are constants

it's defined on functions, that are the limit of characteristic functions (1 on a given set and 0 everywhere else) of measurable sets.
since the rational numbers are measurable with respect to the lebesgue measure, the above function is lebesgue integrable
>That's just a Riemann integral but with horizontal slices instead of vertical right
I never really understood this characterisation to be honest with you.
I think it's better to not try to bring too much geometric interpretation into this, since that won't help you once you start integrating functions over weird spaces that you can't really imagine anyways.

en.wikipedia.org/wiki/Weierstrass_function#Construction

>The characteristic function of a dense subset of R is always discontinuous everywhere.
R is dense in R :^)

Both the set and its complement have to be dense.

>That's just a Riemann integral but with horizontal slices instead of vertical right? I still can't understand how that can integrate that clusterfuck of a function.

Riemann inetgral use the limit of stairs functions, (constant on some intervals) to define the integral.

Lebesgue integral use the limit of "simple functions" (constant on measurable set).

A measurable set is a set that can be obtained from the open sets through countable union, countable intersection, and complement.
(in practice every set is measurable on R, you need the axiom of choice to "build" a non-measurable set)

piecewise functions that have no limits at their endpoints can still be integrated. you just integrate every single point. that is, you take the length of the line at every point. i think that's how it works at least.

He obviously means Lebesgue integrable.

If a function is only made up of isolated points, is it Riemann integrable? I suppose the integral would be also 0 everywhere

Literally every equation is integratible

lmao, if you plug in 1 into the function it gives you a definite value, it doesn't change as you zoom in you dip
iirc in my analysis class we had an example where it was nowhere continuous yet Riemann integrable, and the function was defined on [math][0,1][/math] (may be defined wrong) as
[eqn] f(x)= \left\{
\begin{array}{ll}
0 & x\in \mathbb{R}/ \mathbb{Q} \\ \frac{1}{q} & x=\frac{p}{q} \in\mathbb{Q}\\ \end{array}
\right. [/eqn]

That function is continuous for all irrationals bro. Also, that is a calc1 example, you need ro revise shit.

...

oh shit i would definitely shortsell right there

underrated post

Not an academic but I am a master in logic thinking. Could pic be like a graph but one where lines curves and rotation could be represented

Literally everything function can be integrated.

...

I know the Weierstrass function is nowhere differentiable, but I just thought just now, why can't I simply differentiate term by term in the infinite series? I do that to get non-closed form integrals by using the Taylor series of stuff like e^(x^2), and that's always been accepted, so does the fact that the derivative/integral of sums is the sum of the derivatives/integral fail in infinite sums? If so, why can I integrate e^(x^2) by using the taylor series (or is this not technically legitimaite)?

There are more conditions, in general, such that the derivative of an infinite series is the series of the individual derivatives than the same shit for integrals. I think that for you to "echange" the integral sign with the sum sign you just need uniform convergance of the series, but that doesn't hold for derivatives.

I thought integrating was like sampling a signal - so how is it that some functions cannot be integrated? Is it something similar to aliasing preventing it?

Look at the construction in
If you differentiate each individual term you'll get a factor of (ab)^n. Since ab>1 this series won't converge for any non integer x. So even if you could swap the limits, the result will be bullshit.
Swapping of limits in general is a pretty tricky thing to do.

I bet Weirstrass was chuffed to beans to discover a curve that makes a big W. Now everyone can see at a glance who invented this curve.