Riemann integrability with a discontinuity

[Kolika28]

TL;DR

Given

$f(x) = \begin{cases} 5 & \quad \text{if } x \text{ <3}\\ 7 & \quad \text{if } x \geq3 \end{cases}$

with partitioning $Pn=[0,3−\frac{1}{n},3+\frac{1}{n},4]$ where n∈N and $I=[0,4]$.

Is the function integrable on I?

So, I know that a function is integrable on an interval [a,b] if

$U(f,P_n)-L(f,P_n)<\epsilon$

So I find $U(f,P_n$ and $L(f,P_n$

$L(f,P_n)=5(3-\frac{1}{n}-0)+5(3+\frac{1}{n}-(3-\frac{1}{n}))+7(4-(3+\frac{1}{n}))=22-\frac{2}{n}$
$U(f,P_n)=5(3-\frac{1}{n}-0)+7(3+\frac{1}{n}-(3-\frac{1}{n}))+7(4-(3+\frac{1}{n}))=22+\frac{2}{n}$

Then $U(f,P_n)-L(f,P_n)=\frac{4}{n}$

So I'm struggeling to see if this fraction is less than epsilon or not. The "n" is confussing me. I have done similar tasks, but then I was given a fixed partition expressed with epsilon in the text. Some of my fellow students have tried to explain, but I still don't understand. Does someone have a good explanation?

 

[Gaussian97]

Well, I think your problem is that you don't know what $\varepsilon$ is?
The theorem says the following;
A function$f$ defined and bounded in $[a,b]$ is integrable in $[a,b]$ iff $\forall \varepsilon >0, \exists \Pi$ such that $U(f, \Pi)-L(f, \Pi)<\varepsilon$ where $\Pi$ is a partition.

You now have a set of partitions $P_n$ and you have shown that for all of them $U(f, P_n)-L(f,P_n)=\frac{4}{n}$. Now it's the same as proving the limit of a function. You must prove that, no matter what is the value of $\varepsilon$ you can find a value of $n$ such that $$\frac{4}{n}<\varepsilon$$

 

[member 587159]

For $n$ large enough this becomes smaller than $\epsilon$ (archimedian property).

 

[Kolika28]

Well, I think your problem is that you don't know what $\varepsilon$ is?
The theorem says the following;
A function$f$ defined and bounded in $[a,b]$ is integrable in $[a,b]$ iff $\forall \varepsilon >0, \exists \Pi$ such that $U(f, \Pi)-L(f, \Pi)<\varepsilon$ where $\Pi$ is a partition.

You now have a set of partitions $P_n$ and you have shown that for all of them $U(f, P_n)-L(f,P_n)=\frac{4}{n}$. Now it's the same as proving the limit of a function. You must prove that, no matter what is the value of $\varepsilon$ you can find a value of $n$ such that $$\frac{4}{n}<\varepsilon$$

Okay, I see. So the goal is to prove that there exist a set of partitions that makes this statement true? And then as a result of this, the graph will be integrable on the interval? So it will not be true for every partition? But how do I find this value of $n$ such that $\frac{4}{n}<\varepsilon$
Is it just to say that $n>\frac{4}{\epsilon}$ ?

 

[Gaussian97]

Okay, I see. So the goal is to prove that there exist a set of partitions that makes this statement true? And then as a result of this, the graph will be integrable on the interval? So it will not be true for every partition? But how do I find this value of $n$ such that $\frac{4}{n}<\varepsilon$
Is it just to say that $n>\frac{4}{\epsilon}$ ?

Exact, I like to imagine this as a game, I say an arbitrary value of $\varepsilon$ (for example, 0.7, 0.04 and 0.0035) and you must find a value of $n$ that fulfils the inequality $$\frac{4}{n}<\varepsilon$$In these cases, you can say, for example, n=(6, 287 and 1 151).
But yes essentially you see that doesn't matter what value of $\varepsilon$ I take, any value $$n>\frac{4}{\varepsilon}$$ will fulfil the condition. @Math_QED has given a more formal argument of why you can always find such a number $n$.

 

[Kolika28]

Exact, I like to imagine this as a game, I say an arbitrary value of $\varepsilon$ (for example, 0.7, 0.04 and 0.0035) and you must find a value of $n$ that fulfils the inequality $$\frac{4}{n}<\varepsilon$$In these cases, you can say, for example, n=(6, 287 and 1 151).
But yes essentially you see that doesn't matter what value of $\varepsilon$ I take, any value $$n>\frac{4}{\varepsilon}$$ will fulfil the condition. @Math_QED has given a more formal argument of why you can always find such a number $n$.

Okay, so I can just state that the function is integrable on $I=[0,4]$ by this conclusion? I'm just currious, if I was given a function where the partition $P_n$ was given like in this task (depending on $n$), but the graph was not integrable on an interval $[a,b]$. How could I observe that? Would we not do the operation on the inequality like we did, and find an expression for $n$ also?

 

[Gaussian97]

Well, I think that the typical example is the following, imagine the function
$$f(x)=\begin{cases}1 & \text{ if } x\in \mathbb{Q}\cap[0,1] \\ 0 & \text{ if } \in \bar{\mathbb{Q}}\cap[0,1]\end{cases}$$where $\bar{\mathbb{Q}}$ are the irrational numbers. Then knowing that between any two rational numbers, there is a irrational number and that, between two irrational numbers there is a rational number. You can easily see that, for any partition you look
$$U(f, P_n)=1, \qquad L(f,P_n)=0$$
so for this equation to be integrable you need that $\forall \varepsilon$, $\exists n$ such that
$$U(f,P_n)-L(f,P_n)=1 < \varepsilon$$So, if in this case I give you $\varepsilon=0.5$, try to find any partition fulfiling this condition.

 

[Kolika28]

Ohh, I understand now! I actually got this example in my book. Thank you so much for all the help, I really appreciate it!

Well, I think that the typical example is the following, imagine the function
$$f(x)=\begin{cases}1 & \text{ if } x\in \mathbb{Q}\cap[0,1] \\ 0 & \text{ if } \in \bar{\mathbb{Q}}\cap[0,1]\end{cases}$$where $\bar{\mathbb{Q}}$ are the irrational numbers. Then knowing that between any two rational numbers, there is a irrational number and that, between two irrational numbers there is a rational number. You can easily see that, for any partition you look
$$U(f, P_n)=1, \qquad L(f,P_n)=0$$
so for this equation to be integrable you need that $\forall \varepsilon$, $\exists n$ such that
$$U(f,P_n)-L(f,P_n)=1 < \varepsilon$$So, if in this case I give you $\varepsilon=0.5$, try to find any partition fulfiling this condition.

 

[mathwonk]

Riemann himself proved, in the same paper where he defined his notion of integration, that a bounded function is integrable in his sense if and only if, for every positive d, the set of points where the function oscillates by d or more, can be covered by a finite sequence of intervals whose total length is as small as you wish. In particular a function has oscillation zero at every point where it is continuous. Your function is discontinuous at exactly one point, namely x=3, where the oscillation is 2. Since it is easy to cover one point by an interval as short as you wish, your function is integrable. It follows also that a bounded function which is discontinuous at only a finite number of points, or even only at an infinite sequence of points, is always integrable. There also exist integrable functions which have an uncountable set of discontinuities.

This condition is equivalent to the later one, stated by Lebesgue, that a bounded function is integrable in Riemann's sense, if and only if the set of points where it is discontinuous can be covered by an infinite sequence of intervals, whose total length is as small as you wish. Since Lebesgue's result is famous, but most people have not read Riemann's paper in detail, this condition is usually attributed, incorrectly in my opinion, to Lebesgue. I.e. Lebesgue's condition is merely an equivalent restatement of Riemann's condition.

 

[zinq]

The Riemann integral is defined to be the limit of the sum of the areas of rectangles whose bases are small intervals and whose heights are any value of the function in that interval ... as the mesh of the partition (the largest size of any of its intervals) approaches 0 ... if that limit is well-defined.

When the mesh is small enough, the interval containing the discontinuity will be very tiny, so the area of its rectangle will be less then 7 * mesh. Which for a small enough mesh can be made as small as you like. (Or, if x = 3 is on the boundary between two intervals of the partition, you can see that the areas of the two adjacent rectangles will add up to 7 * mesh + 5 * mesh = 12 * mesh — which once again is as small as you like when the mesh is sufficiently small. So: The interval or intervals containing the discontinuity at x = 3 can be *entirely ignored* since in the limit they will contribute nothing to the integral.

Which means that the integral is the same as the integral of f(x) = 5 from x = 0 to x = 3, plus the integral of f(x) = 7 from x = 3 to x = 4. Since these are both constant functions, these integrals exist (meaning: their limits exist). And so the original integral exists.

 

[mathwonk]

very good! this shows that the OP could have used the theorem that a function which is integrable on both intervals [a,b] and [b,c] is also integrable on [a,c]. Of course he would have to believe that the function with value 5 on [0,3) and value 7 at x=3, is integrable on [0,3], which uses your argument.