Locally integrable function Puns

I'm attending a signal processing course at my university.

The professor defined a distribution as a functional being both linear and continuous.

He also showed the following theorem:

Hp: [; g(t) ;] is a locally integrable function Th: [; T_g(f) = \int_{-\infty}^{\infty} g(t) f(t) dt ;]

As far as I understand, a locally integrable function is a function whose integral is finite for every closed and limited interval of its domain.

For instance [; \frac{1}{x} ;] is locally integrable, because while [; \int_{1}^{\infty} \frac{1}{x} dt ;] is infinity, we have that [; \int_{a}^{b} \frac{1}{x} dt = log(b) - log(a) ;] with [; [a, b] \in [1, +\infty) ;].

So far so good.

Then he proceeds to show a property of the derivative of a distribution:

Hp: [; T_g(f) ;] is a distribution Th: [; T'_g(f) = -T_g(f') ;]

He demonstrates this using the product rule for derivatives:

[; \frac{d}{dt} g(t) f(t) dt = g'(t)f(t) + g(t) f'(t);]

[; g'(t) f(t) = \frac{d}{dt} g(t) f(t) dt - g(t) f'(t);]

[; \int_{-\infty}^{\infty} g'(t) f(t) dt = \int_{-\infty}^{\infty} \frac{d}{dt} g(t) f(t) dt - \int_{-\infty}^{\infty} g(t) f'(t) dt;]

[; \int_{-\infty}^{\infty} g'(t) f(t) dt = \[g(t) f(t)\]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} g(t) f'(t) dt;]

He says that because [; g(t) ;] is locally integrable then \[g(t) f(t)\]_{-\infty}^{\infty} goes to 0 at both [; -\infty;] and [; +\infty ;].

Why does this happen? Why does it have to go to zero since it is locally integrable?

Thank you in advance.

LateX version for convenience here.

Write L^1 (respectively L^(∞)) for the set of integrable (respectively essentially bounded) functions on [0, 1], with respect to the usual Lebesgue measure, denoted mu.

Given a non null measurable subset A of [0, 1], define the measure space (A, F_A, nu_A) where F_A is the induced sigma algebra, and nu_A is the measure given by nu_A (E) = mu(E n A)/mu(A).

Given a L^1 function f, and a subset A as above, define f_A to be the function on (A, F_A, nu_A) given by f_A (x) = f(x)/mu(A).

Let H be a family of non null measurable subsets of [0, 1]. Find necessary and sufficient conditions on H such that for every f in L^1 the following statement holds:

f is in L^ ∞ if and only if the family {f_A} (A in H) is uniformly integrable.

Title says it all. Does there exist such a function?

Hi everyone so I was searching the internet for conditions that make a function integrable. Here's what I found:

If a function is continuous on a given interval, it’s integrable on that interval. Additionally, if a function has only a finite number of some kinds of discontinuities on an interval, it’s also integrable on that interval.

So does that mean despite having a point of discontinuity in an interval, is the function still integrable on that interval? Based on my understanding, a function can be integrable despite having discontinuities (finite or infinite) as long the interval does not include the discontinuities. Now I'm a bit confused as to how the above statement is correct.

Thank you for answering!

Fourier series allow us to represent any integrable periodic function as an infinite sum of sine and/or cosine functions. My understanding is that all periodic functions over the complex numbers form an infinite dimensional vector space and Fourier series are simply a convenient choice of basis. Am I correct? If not, what am I missing?

If so, does that mean that we could, for example, represent an arbitrary periodic function as an infinite series of square waves? Or triangle waves? Or, in other words, given some specific type of periodic function over the complex numbers, call it T, is it possible to represent every other function in the aforementioned vector space as an infinite series of functions of type T?

I really hope he doesn’t use reddit, but last night our roommate came home late- VERY DRUNK. By 3 AM, me, along with all two of my other roommates were forced to witness him attempt to converge Cauchy sequences all over his bed. Worst of all, he was too drunk to even realize at the moment. None of us were courageous enough to approach the mess, so it stayed like that all night, until he woke up and used L^p spaces instead. However, the long night left this unbounded mess of linear operators. We took finite-rank truncations of them but it still doesn’t cut it. Now it’s getting late and I’m writing this balls deep in Stein/Shakarchi. I wouldn’t dare go back in that quad with that nonconstructive Vitali function.

The Function,

f(x) = 1, if x is rational

= 0, if x is irrational

has no reimann integral over [0,1] i.e the function is not integrable over this interval.

can anyone explain this?how and why.

Hello guys. I'm studying a theory of integration (of Henstock-Kurzweil) which extends the class of integrable functions with R as domain. Its main feature is an extension of the fundamental theorem of calculus, which states that if a function f has a primitive, then f is integrable without further hypothesis on f; to set it apart from lebesgue, I'm looking for an example of a function (defined on a finite interval) which has a primitive on the interval but that is not Lebesgue-integrable (the existence of such function is hinted at in many issues I've read, yet I didn't manage to find an explicit one). I figured it must be something that is not absolutely integrable (that's basically what sets the Henstock-Kurzweil integral apart from the Lebesgue integral), so I think it could be some variant of (1\x)*sin(1\x) on [0,1], but I didn't have any luck yet.

EDIT: I just realized I said twice "finite interval" instead of "bounded interval", sorry xD

So I've got a bunch of exercises for homework that tell me to prove that certain piecewise functions are integrable on an interval. We got taught the Lebesgue theorem (if a function has a finite amount of discontinuity points on an interval then it's continuous) as well as a theorem which states that if a function is continuous on an interval then it's integrable on that interval. I guess that second theorem would sort of be a particular case of the Lebesgue theorem since a continuous function has 0 discontinuity points, and 0 is finite, but that's not important.

In class we solved all these functions by calculating the lateral limits and seeing whether the function is continuous, or, if it's not, what are the discontinuity points equal to. So take for example the function

f : [-2; 2] -> R, f(x) = {

3x+2, if x is part of [-2; 1)

4x^(2)-2x+1, if x is part of [1; 2]}

So we'd start by saying that f is continuous on [-2; 2] \ {1} because on that interval it is a polynomial function (1) and then calculate limit from the left and from the right in 1 which are 5 and 3 respectively, which means that the function has one discontinuity point (2) and from (1) and (2) it means that the function is integrable on [-2; 2] by the Lebesgue theorem.

What I just realized is that I could cheat my way in all of these exercises and not need to calculate the lateral limits. Proving that it's continuous on [-2; 2] \ {1} (the more general case: on its domain other than a finite set of points) is enough, since if the lateral limits in 1 were equal, then the function would be continuous and, thus, integrable. If they were not equal then the function would have a single (a finite amount) of discontinuity points. This sort of feels like cheating however since now I could do an 1h homework in 5 minutes without needing to calculate any limits, but it seems correct to me.

So is it enough to say that a function is continuous on its domain minus a finite amount of points to prove that it's integrable?

I'm working through Stein and Shakarchi's Princeton Series in Analysis, starting with their first book on Fourier analysis. I'm looking at a theorem (Theorem 4.1 in Chapter 2.4, pg. 49 in the book I'm using) for the convergence of a certain convolution, between a function f(x) satisfying:

[Riemann] integrable on the circle,

and a "good kernel" which places all of its weight at x = x0 in the limit. The value of the convolution, defined only for a 2π-periodic interval, approaches f(x0) in the limit, so long as f(x) is continuous at x0. (The theorem also states that if f(x) is continuous everywhere in the closed interval, then the limit is uniform.)

I'm fine with the idea behind the theorem, but a step in the proof for the first part asserts that f(x) is bounded within the closed interval, say [-π, π]. Integrability does not itself imply boundedness, but am I correct that being defined on the closed interval does? It's not made explicit that f(x) returns finite values for all x in [-π, π], though...

Hello everyone.

I have the following function:

> f(x) = \frac{\sin(x) \cdot e^{x}}{x}

And I'm supposed to show that it's integrable in the [0,1] interval, then verify the inequality of:

> 0 \leq \int_0^1 f(x) dx \leq e-1

I don't really know how to do this. My plan was to try out if the function was Riemann integrable by comparing the upper sum and the lower sum.

But my upper sum:

> U(P_n,f) = \frac{1}{n} = \frac{1}{n}[f(x_1) + f(x_2) + ... + f(x_n)]

Then ends up being:

> \frac{1}{n} [ \frac{\sin(x_1) \cdot e^{x_1}}{x_1} + \frac{\sin(x_2) \cdot e^{x_2}}{x_2} + ... + \frac{\sin(x_n) \cdot e^{x_n}}{x_n}

And I have no clue how to go from there.

Wouldn't it just end up looking like this? I'm very stuck here :(

And also not sure how to verify the inequality that is asked for in the second part.

If anyone could help me out, I'd really appreciate it!

Hello, as I was reading my teacher's note I stumbled upon the definition of a Lebesgue-Integrable function:

Given the measured space 𝛺 = ( 𝛺, A, 𝜇 ) and the measurable function f : 𝛺 -> ℝ ∪ { ± ∞ }, we say that f is Lebesgue-integrable if ∫ |f| is finite.

So my question is: Isn't it too much to extend the definition to include functions in the extended real line ? I can't find an example where we need it. So is there a function that is Lebesgue integrable on a set A such that f(A) ⊄ ℝ ?

Thank you :)

Edit: Yeah I found out we can construct such a function without much work. Let A ⊂ ℝ such that it is countable. We can define f : ℝ -> ℝ ∪ { ± ∞ } with x ↦ 𝜒_A(x) * ∞ where 𝜒_A is the indicator function of A. Clearly it is well define (0 * ∞ = 0 by convention in my measure theory class), and also f(A) = {0, ∞} ⊄ ℝ but since A is countable and Lebesgue-negligable, f = 0 almost everywere so it Lebesgue-Integral has value 0.

Hi!
i can’t figure out the answer to this math problem (link below).

can anyone help me?

https://imgur.com/a/YawhiP5 (Exercise’s Text)

https://imgur.com/a/8OBCG47 (How i done it)

Sorry for bad English

Thanks in advance

Hey group,

I am new to Google Cloud Functions and have been tasked with refactoring some for my job.

Problem:

• I want to trigger a series of cloud functions locally for test purposes

Attempted Solution:

• I installed the functions_framework and have attempted to run it in terminal but am getting an error.
• I have imported functions_framework into my main.py module
• I'm using the <@functions_framework.http> annotation on the function I want to trigger

Error:

• OSError: Project was not passed and could not be determined from the environment.

Thoughts:

• I'm wondering if I need to config gcloud to work with a given project.

Help appreciated. Thanks

We learnt that a |mod| function is continuous yet non-differentiable and as integrals are defined as "anti-derivatives" sooo how come we can integrate a mod function yet we can't differentiate it?

Hi, all. I'm preparing a presentation on the Conway base 13 function (Wikipedia article here) for my Real Analysis class, and I'm trying to nail down one final detail: Is this function integrable?

Intuitively, the answer feels like no to me, but I can't quite justify why. I know that it being discontinuous does not imply that it is not integrable, but it just feels impossible with how strange the function is.

There's also a part of me, however, that feels like the integral would be zero. If the function is generally zero but also maps to every real number, wouldn't the positive real numbers cancel out with the negative real numbers, resulting in zero?

Edit: Thank you for your replies! Sorry, I should have clarified. I'm referring to Riemann integration here, as we haven't learned about Lebesgue integration.

Hi, I need help with this problem.

I'm confused with how Riemann sums work on double integrals. I know that L=∑_i,j m_ijA_ij and U=∑_i,j M_ijA_ij where m_ij is the greatest lower bound and M_ij is the least uper bound and A_ij is the area of each partition.

A_ij=1/n^(2) for every partition. If I get it right, M_ij must be where x_i and y_i are the maximum value on the square, so it must be at the the right top corner, while m_ij must be at the bottom left corner. So M_ij=(i/n,j/n) and m_ij=((i−1)/n,(j−1)/n, right?

I don't know that to do next, because of the piecewise behavior. Hope you can help me.

Exercise: Let f be a measurable function from R^d to C that is supported on a set of finite measure and let ε>0. Show that there exists a measurable set E of measure at most ε such that f is locally bounded outside of E.

Solution attempt: Let n be a positive integer. Then there is a compactly supported continuous function g such that (the L^1 norm) ||f-g||< ε/2^n . The set E(n)={x in B(0,n): |f(x)-g(x)|>1} has measure ∫_{E(n)} 1 leq ∫_{E(n)} |f-g| < ε/2^n . Since g is continuous on the closed ball B(0,n) and this ball is compact, so there is some real number M(n) such that |g(x)|<M(n) for all x in B(0,n). If x is not in E(n) then |f(x)| leq |g(x)|+1 <M(n)+1.

If E is the union of all E(n), then its measure is less than ε. Given any positive real number r, let R denote the next integer larger than r. The map f is locally bounded outside of E because |f(x)|<max{M(1),…,M(R)}+1 for all x outside of E.

Is this a correct solution? If not, where is the problem?

I have function that has a loop. In each iteration of the loop, it asks the user for input and adds that input to a larger string. So with each iteration, the string gets bigger. After the loop is over, the function returns that string.

I have done a print statement within that function and I can confirm the string is what I want it to be.

However in my main function, when I declare a new variable salad by assigning it that function, salad is blank. But the combo string inside the function has the fruits.

Why is this happening?

My guess is, when I write that assignment statement in the main function, the salad variable is immediately getting assigned with an empty string (since that's what the combo string starts out as)? How do I make it return the string only after the loop is over?

def main():
    salad = makeSalad()    





def makeSalad():
    combo="";    

    for x in range(0, 4);
    ingredient = input("Please enter the ingredient");
    combo = combo + ingredient;

return combo;

u(t,x) = 1/x^2 * sin^2(x) exp(-tx) over (0,\infty) with t \ge 0

1. with |sin^2 (x)/x^2| \le 1, so i have to integrate over exp(-tx) i can solve the integral to 1/t - exp(-t)/t over [1,\infty)

2. but for (0,1) i have no clue. the problem is really that 1/x^2 goes besark when x ->0, sin^2(x) goes to 0 and exp to 1.

i tried to expand exp and sin^2 in series and see if something vanishes, but this also leads to nothing, which i can use. :(

u(t,x) is continous over x>0 ==> borel measurable, first part in the definition of \mathscr{L}^1 (\lambda), would require to integrate int_1^ {\infty} |u(t,x)| d\lambda < \infty

not really a fan of this function.

Define f(x) in this way: Let rational x's =1 and irrationals = 0 on the interval [-1,1]. I argue that f(x) is not integrable because it is discontinuous everywhere using an application of the limit definition. I know that all continuous functions are integrable but the inverse isn't necessarily true so I have a feeling my answer is wrong.

How would I prove that a function that has many of these discontinuities is still integrable?

As the title says, I am having a tough time working through proofs of Integration in my Real Analysis class and would like to better understand the topic in general. Where do I start to really solidify this topic in my head?

Specifically: I give you a set of functions f1(x), f2(x), etc. all the way to n (or infinite). All of them can be integrated over some interval I. I ask if there's another function g, such that the integral over I of fi(x) fj(x) g(x)dx is nonzero if and only if i = j?

For example, let's say that f1(x) = 1, f2(x) = cos(x), and f3(x) = 1/2(3cos(x)^2 -1), over the interval 0 to pi. Then g(x) = sin(x) works. (those are the legendre polynomials).

How do we go about finding a g(x), if it always exists? This was inspired by my physics textbook, in which we use Fouriers trick to find the potential of a charge distribution.

I have your standard programmable thermostat for my gas furnace and AC system. I think it's time I brought my HVAC under HA control.

How much pain am I in for? What's your experience?

In quantum mechanics, square-integrable functions are often encountered as wavefunctions. By square-integrable it just means that the function reduces to 0 at infinity. Schwartz functions share this property, but apparently have some stricter conditions. Is there a distinction between the two, or can the terms be used interchangably?

Hi!

The exercise is as follows:

We have a function f:[0,1]--> R with 0<= f(x) <= 1 for all x in [0,1]. Also, for every p>0 the set A_p := {x in [0,1] | f(x) > p} is finite.

I'm not sure if I have written the following proofs properly. So any advice with that would be appreciated.

a.) Show Riemann-under integral f(x) dx >= 0.

Because U(f,P) <= Riemann-under integral and U(f,P) = sum(j=1 to n) sup f * (x_{j} - x_{j-1}) >= 0 for all partitions (and interval length >0 per definiton).

Let p>0 be given randomly and P = {x0, x1, x2, ..., xn}.

b.) Prove that at the most 2| A_p | indices of j have sup(I(j)) f > p.

Because each interval is closed, there exist points that are contained in two intervals. And because | A_P | is the amount of elements that have f(x) > p we find the desired expression.

c.) Prove that U(f,P) <= 2m| A_p | + p, where m = max(1<= j <= n) (x_{j} - x_{j-1}), I think this is called measure in English?

We have U(f,P) = sum(j=1 to n) sup f * (x_{j} - x_{j-1}) <= sum(j=1 to n) sup f * m <= 2m| A_p | < 2m| A_p | + p.

I'm not sure if the second inequality is justified enough, because we can say sup f <= 1, then it follow immediately? The last statement doesn't make sense to me, is it a more inaccurate bound? I don't see why this would be useful later on.

d.) Prove the function f is Riemann integrable and prove integral(0 to 1) f(x) dx = 0

We see that f is bounded on its domain, namely |f(x)|<=1. Then a function is Riemann integrable if and only if for every epsilon>0 there exists a partition such that U(f,P) - L(f,P) < epsilon.

I'm not sure how to bound L(f,p). Intuitively, I think that L(f,P) is 0 everywhere, except for a finite amount of discontinuities.

How can I prove L(f,P) = 0? My attempt:

We prove by contradiction. Assume L(f,P) =/= 0, then there exists an interval with index J where inf(I(j)) > 0. This is a contradiction with the property that says there are finite amount of x such that f(x) > p because the interval has infinite points.

Choose m< p/(2 | A_p |), then we have U(f,P) - L(f,P) <= 2m| A_p | + p < 2p := epsilon. We conclude f is Riemann integrable on I=[0,1].

It feels like this proof is incomplete. We can choose m however we like, right? Because m>0, there are finite number of intervals and p>0 is given randomly, so we can say 2p := epsilon?

e.) New function: g:[0, 1] --> R, defined by g(x)=0 if x in [0,1]\Q and g(x) = 1

Wikipedia's article on Lebesgue integration seems to hint that these integrals have some practical importance, particularly with Fourier transforms, but doesn't really give any practical examples. This article seems to suggest that maybe there aren't too many real world applications of Lebesgue integrals. Not to say that they aren't interesting and important from the pure math side, but I would be interested if there was anything from say physics to also motivate them.

The Function,

f(x) = 1, if x is rational

= 0, if x is irrational

has no reimann integral over [0,1] i.e the function is not integrable over this interval.

can anyone explain this?how and why.

How can I prove that

5cos(y)/(x+1) if y<=x

x+3 if y>x

is integrable over the rectangle [0,3]x[0,2]?