I don't know any reference, but the construction can be done as follows. Consider the space $\ell_p$ for $p = 1/4$. This space equipped with distance $\rho(x, y) = \|x - y\|^p$ is not locally convex complete metrizable topological vector space. Denote $e_n$ the vectors of the canonical basis of $\ell_p$, that is $e_1 = (1, 0, 0, ...)$, $e_2 = (0, 1, 0, 0, ...)$, etc. Let $\Delta_n = [2/(2n+1), 1/n]$. The intervals $\Delta_n$ are mutually disjoint and tend to 0. Now define the required function $f: [0, 1] \to \ell_p$ as follows: on every $\Delta_n$ the function $f$ takes the constant value $e_n/n$; between two neighboring intervals $\Delta_n$, $\Delta_{n+1}$ define $f$ by means of linear interpolation (in order to make it continuous), and put $f(0) = 0$.

Let us demonstrate that this continuous function $f$ is not integrable. Indeed, if $x = (x_1, x_2, ...) \in \ell_p$ is the integral of $f$, then for each $n$, applying $n$-th coordinate functional we have that $x_n \ge \int_{\Delta_n} 1/n dt = \frac{1}{n^2(2n+1)}$. Consequently,

$$

\|x\|^p = \sum_n |x_n|^p \ge \sum_n \left( \frac{1}{n^2(2n+1)} \right)^{1/4} = \infty.

$$

This contradiction completes the proof.

Comment posted on May 22, 2018

I thank Vladimir Kadets a lot for his wonderful example.

Is it correct that “Integrable” means”Iintegrable with respect to Pettis integral”?

https://en.wikipedia.org/wiki/Pettis_integral

Below are some additional explanations for this example to help the readers.

This is a first draft. I need a little more time to recheck everything. I am totally overflowed.

The example is a function f from [0, 1] into l^p, with p = ¼.

In the terrible case when 0 < p < 1, the spaces l^p are defined on the following Web page:

https://en.wikipedia.org/wiki/Sequence_space

Recall that every element of l^p is a sequence (a1, a2, a3, …) of real numbers.

Recall that we can multiply such a sequence by a real number r:

r (a1, a2, a3, …) = (ra1, ra2, ra3, …)

= (a1 times r, a2 times r, a3 times r, …) = (a1, a2, a3, …)r.

If r is not zero, we have (a1, a2, a3, …)/r = (a1, a2, a3, …)(1/r)

= (a1(1/r), a2(1/r), a3(1/r) …) = (a1/r, a2/r, a3/r …).

The elements of the canonical basis of l^p are denoted by e(1), e(2), e(3), …, that is:

e(1) = (1, 0, 0, 0, 0, … ),

e(2) = (0, 1, 0, 0, 0, … ),

e(3) = (0, 0, 1, 0, 0, … ), ….

First, we divide the interval [0, 1] into subintervals.

Note that 0 < … 2/5 < 2/4 < 2/3 < 2/2 = 1.

We define

I(1) = [2/3, 2/2] = [2/3, 1],

I(2) = [2/4, 2/3],

I(3) = [2/5, 2/4],

…,

and generally, for every k in {1, 2, 3, …},

we define the subintervals I(k) of [0, 1] by

I(k) = [2/(2k + 2), 2/(2k + 1)].

Since … 2/5 < 2/4 < 2/3 < 2/2 = 1,

and [limit of 2/n when n goes to infinity] = 0, we have

]0, 1] = … union [2/5, 2/4] union [2/4, 2/3] union [2/3, 2/2], so that

]0, 1] = … union I(3) union I(2) union I(1).

Now, we start constructing the function f from [0, 1] into l^p

so that for every x in [0, 1], the value f(x) will be an element of l^p.

In particular,

for every x in [0, 1], the value f(x) will be a sequence of real numbers.

Now we define f as a fixed sequence on each of the subintervals I(1), I(3), I(5), …:

For every x in I(1), we define f(x) = e(1)/1 =(1, 0, 0, 0, …).

For every x in I(3), we define f(x) = e(2)/2 = (0, 1/2, 0, 0, …).

For every x in I(5), we define f(x) = e(3)/3 = (0, 0, 1/3, 0, …).

Generally, for every k in {1, 2, 3, …}, for every x in I(2k - 1), we define f(x) = e(k)/k = constant.

So far we have defined f on … union I(5) union I(3) union I(1),

that is, we have defined f on

… union [2/7, 2/6] union [2/5, 2/4] union [2/3, 2/2].

Now we define f on the subintervals I(2), I(4), I(6), …,

by linear interpolation, so that f will be continuous on ]0, 1].

On I(2) = [2/4, 2/3], we will interpolate linearly from f(2/4) to f(2/3).

On I(4) = [2/6, 2/5], we will interpolate linearly from f(2/6) to f(2/5)

On I(6) = [2/8, 2/7], we will interpolate linearly from f(2/8) to f(2/7). ….

Generally, for every k in {1, 2, 3, …},

on I(2k) = [2/(2k +2), 2/(2k +1)],

we will interpolate from f(2/(2k +2)) to f(2/(2k +1)).

Recall that given two vectors A and B in a vector space V,

to define a function g from a nontrivial interval [a, b] into V

by linear interpolation such that g(a) = A and g(b) = B, we define for every t in [0, 1],

g((1 – t)a + tb) = (1 – t)A + tB,

which gives

when t = 0, g(a) = A as wanted,

and when t = 1, g(b) = B as wanted.

For every t in [0, 1], let x = (1 – t)a + tb = a + t(b – a),

so that x – a = t(b – a),

and t = (x – a)/(b – a).

Now, we can define f on the subintervals I(2), I(4), I(6), … by this linear interpolation.

Let k in {1, 2, 3, …}.

On I(2k) = [2/(2k +2), 2/(2k +1)], we will interpolate

from f(2/(2k +2)) = e(k + 1)/(k + 1)

to f(2/(2k +1)) = e(k)/k.

For this, we apply our above formula with

a = 2/(2k +2),

b = 2/(2k +1),

A = e(k + 1)/(k + 1),

B = e(k)/k.

Let x be any point in the subinterval I(2K) of [0, 1].

Let t = (x – a)/(b – a). Then we define f(x) = g(x) = (1 – t)A + tB.

Now, we have defined f on ]0, 1].

Since ]0, 1] = … union I(3) union I(2) union I(1),

and f is constant on the subintervals I(1), I(3), I(5), …,

and f is defined by linear interpolation on the subintervals I(2), I(4), I(6), …

we obtain that f is continuous on ]0, 1].

Now we define f(0) = 0.

Since e(k)/k goes to zero when k goes to infinity, we have that f(x) goes to 0 when x in ]0, 1] goes to 0.

It follows that f is continuous on [0, 1].

Let us prove by contradiction that f is not integrable on [0, 1].

Suppose that f is integrable.

For any function h from an interval J of R into a topological vector space, let S(h, J) denote the integral of h over J.

By construction, f is a sequence (f1, f2, f3, …) of positive real numbers.

For every k in {1, 2, 3, …}, we have for every x in I(2k - 1), f(x) = e(k)/k, so that fk(x) = 1/k

From the properties of Pettis integrals, we have S(f, [0, 1]) = (S(f1, [0, 1]), S(f2, [0, 1]), S(f3, [0, 1]), …).

For every k in {1, 2, 3, …}, we have for every x in I(2k - 1), f(x) = e(k)/k, so that fk(x) = 1/k

S(fk, [0, 1]) > S(fk, I(2k -1)) = (1/k)[Length of I(2k – 1)]

Length of I(2k – 1) = 1/(2k – 2) – 1/(2k – 1) = 1/(2k – 2)(2k – 1),

S(fk, [0 ,1]) > (1/k)[1/(2k – 2)(2k – 1)].

The end of the proof is essentially the same as what Vladimir has written.

I need to recheck everything carefully.

I hope that my first draft above helps.

With best regards, Jean-Claude

Every continuous function is Riemann integrable, and every Riemann integrable function is Lebesgue integrable, so the answer is no, there are no such examples.

Let $g(0)=1$ and $g(x)=0$ for all $x\ne 0$. It is straightforward from the definition of the Riemann integral to prove that $g$ is integrable over any interval, however, $g$ is clearly not continuous.

The conditions of continuity and integrability are very different in flavour. Continuity is something that is extremely sensitive to local and small changes. It's enough to change the value of a continuous function at just one point and it is no longer continuous. Integrability on the other hand is a very robust property. If you make finitely many changes to a function that was integrable, then the new function is still integrable and has the same integral. That is why it is very easy to construct integrable functions that are not continuous.

Probably the simplest example of an integrable function that's not continuous is something like $$f(x)=\left\{\begin{array}{rl}3 & 0\leq x

Answer posted on May 23, 2018

Dear Rafik Zeraoulia,

Two days ago, Vladimir Kadets posted an example of a continuous function from [0, 1] into a topological vector space that is not integrable. This illuminating, wonderful, striking example is posted on this page above your answer. I think that it is the simplest possible example. I will update the information on my Research Gate project about this as soon as possible. Concerning continuous functions from [a, b] into a topological vector space, I must split my project “Non-locally convex spaces” into at least two projects: A research project with information about “Continuous on [a, b], not integrable” and another research project with information about “Derivative zero and not constant”.

With best regards, Jean-Claude

Any continuous function is Lebesgue measurable on closed interval [a,b].

Answer posted on May 23, 2018

Dear Amirhossein Sobhani, the concept of “Lebesgue measurable function” is defined for functions with values in the set of complex numbers, not for functions with values in a topological vector space different from the set of complex numbers.

https://en.wikipedia.org/wiki/Measurable_function#Notable_classes_of_measurable_functions

With best regards, Jean-Claude

Continuity ln a closed bounded interval implies Riemann integrability because an upper sum approaches a lower sum as the refinements become finer.

There seems to be two questions here. One is is a continuous real valued or complex valued function on a closed bounded interval of the real line integrable. As many have pointed out - you don't even need measure theory for that since it is Riemann integrable. The second question is a function from the same interval into a topological vector space. Here we are taking about a vector valued integrable defined on that vector space and not the normal Lebesgue integral for complex valued function or the Boehner integral for functions into a Banach space which have similar properties to the Lebesgue integral except for possibly the Radon-Nikodym property which is true for reflexive Banach spaces and of course Hilbert spaces.

As pointed the answer to this question depends on the topological properties of the range space.

Answer posted on May 24, 2018

I thank Thomas Carter and Truman Prevatt for their answers.

======================================

I recall that three days ago, Vladimir Kadets posted an example of a continuous function from [0, 1] into a topological vector space that is not integrable.

This illuminating, wonderful, striking example is posted above on this page.

It is the first example answering the original question.

I think that it is the simplest possible example answering the original question

======================================

As Truman said, Bochner integrals are integrals of functions from a measure space into a Banach space:

https://en.wikipedia.org/wiki/Bochner_integral

https://en.wikipedia.org/wiki/Salomon_Bochner

======================================

Integrals from a measure space into a topological vector space are called “Pettis integrals” or “Gelfand-Pettis integrals:

https://en.wikipedia.org/wiki/Pettis_integral

https://en.wikipedia.org/wiki/Billy_James_Pettis

https://en.wikipedia.org/wiki/Israel_Gelfand

======================================

It is interesting to take the opportunity of the original question to add the following instructive pieces of information mentioned by Truman:

1. The Radon–Nikodym theorem fails to hold in general for Bochner integrals from a measure space into a Banach space.

2. This theorem holds in the special case where the Banach space is reflexive.

3. In particular, it holds when the Banach space is a Hilbert space.

https://en.wikipedia.org/wiki/Bochner_integral#Radon–Nikodym_property

https://en.wikipedia.org/wiki/Bochner_integral#See_also

https://en.wikipedia.org/wiki/Radon–Nikodym_theorem

======================================

Salomon Bochner created the Bochner integral in 1933:

https://en.wikipedia.org/wiki/Salomon_Bochner#Mathematical_work

======================================

With best regards, Jean-Claude 

Dear Jean-Claude,

I only would like to remark, that when I was speaking about the integral of functions acting from an interval to a non-locally convex space $X$, I meant the Riemann integral. It can be defined basically the same way as for scalar-valued functions: first, for every partition $\{\Delta_k\}_{k=1}^n$ of the domain and every choice of $t_i \in \Delta_i$ define the integral sum

$$

\sum_{k=1}^n f(t_k) |\Delta_k| \in X,

$$

and then define the integral as the limit of integral sums.

The Pettis integral that you mention, is not applicable for general non-locally convex spaces, because this definition uses linear functionals, but some "bad" non-locally convex spaces do not possess continuous linear functionals (outside of zero functional) at all. Nevertheless, in non-locally convex $\ell_p$ there are continuous linear functionals, and what I am using is just the easy fact that for a continuous linear functional $g$ on $X$, and for a Riemann integrable function $f: [0, 1] \to X$ one has the formula

$$

g\left(\int_[0,1] f(t) dt\right) = \int_[0,1] g(f(t)) dt.

$$

Best, Vladimir

Answer posted on May 24, 2018

I thank Vladimir Kadets a lot for his huge clarification for working in the minefield of non-locally convex topological vector spaces.

Thanks to his huge clarification, I have just found what seems to be the most important publication of my life in mathematics. It is the following:

======================

INTEGRAL REPRESENTATION THEOREMS IN TOPOLOGICAL VECTOR SPACES

ALAN H. SHUCHAT

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY

Volume 172,

October 1972

http://www.ams.org/journals/tran/1972-172-00/S0002-9947-1972-0312264-0/S0002-9947-1972-0312264-0.pdf

======================

I am going to eat a copy of this fantastic publication for my lunch.

With a lot of thanks to Vladimir, and with best regards, Jean-Claude

no there is not all all continuous function are itegrable

Answer posted on May 29, 2018

Dear Louiza derbal,

About 8 days ago, as the first answer above on this page, Vladimir Kadets posted a wonderful simplest possible example of a function defined on the interval [0, 1] of the set R of all real numbers with values in a topological space that is continuous and not integrable. About 5 days ago, above on this page, Vladimir specified that the considered integrability is the ordinary Riemann integrability. The following links give direct access to his two postings:

https://www.researchgate.net/post/Are_there_examples_of_continuous_functions_defined_on_a_b_that_are_not_integrable#view=5b032c6fc1c6b186a0102abb

https://www.researchgate.net/post/Are_there_examples_of_continuous_functions_defined_on_a_b_that_are_not_integrable#view=5b06df2b6a21ff47cc0e188e

Or go to the following Web page, and then to “Contributions”, then to “Answers”, and click on “View” not on the title of the answers:

https://www.researchgate.net/profile/Vladimir_Kadets

With best regards, Jean-Claude

Comment posted on May 29, 2018

==========================================

The only example that has been posted on this page of a continuous and not integrable function defined on a non-empty closed interval of the set R of all real numbers is the major example posted by Vladimir Kadets.

It is the first answer posted on this page.

The following links gives direct access to Vladimir’s example:

https://www.researchgate.net/post/Are_there_examples_of_continuous_functions_defined_on_a_b_that_are_not_integrable#view=5b032c6fc1c6b186a0102abb

https://www.researchgate.net/post/Are_there_examples_of_continuous_functions_defined_on_a_b_that_are_not_integrable#view=5b06df2b6a21ff47cc0e188e

==========================================

In this example, the topological space is the space l^p, with p = ¼.

https://en.wikipedia.org/wiki/Sequence_space

This space is a metric space, not a normed space.

==========================================

The following article gives a connection between “Metric space” and “Generalized normed space”:

A natural Galois connection between generalized norms and metrics

Arpad Szaz

Acta universitatis sapientiae. Mathematica 9(2):360-373

DOI: 10.1515/ausm-2017-0027

December 2017

https://www.researchgate.net/publication/324911700_A_natural_Galois_connection_between_generalized_norms_and_metrics

==========================================

With best regards, Jean-Claude prelo

Answer posted on June 1, 2018

Dear Benaissa Bouharket,

On May 21, 2018, Vladimir Kadets posted a striking simplest possible example of a function defined on the interval [0, 1] that is continuous and not integrable. This function takes its values in a topological space. This example is located on the following part of this page:

https://www.researchgate.net/post/Are_there_examples_of_continuous_functions_defined_on_a_b_that_are_not_integrable#view=5b032c6fc1c6b186a0102abb

On May 24, 2018, Vladimir specified that the considered integrability is the ordinary Riemann integrability. This piece of information is located on the following part of this page:

https://www.researchgate.net/post/Are_there_examples_of_continuous_functions_defined_on_a_b_that_are_not_integrable#view=5b06df2b6a21ff47cc0e188e

With best regards, Jean-Claude

Continuity of a function on [a,b] \in R is one of the sufficient conditions for Riemann Integrability. So, the answer is NO.

It is not possible in metric spaces. You can try with (non-metrizable) Topological Spaces

f(x)=tan x, 0

I agree with Parameshwari Kattel