"Imagine" a definite integral with a finite number of discontinuties

[dRic2]

Hi, I'm re-studying integrals and I got stuck with this problem. Actually the math beyond it is very clear but I still can figure it out.

Take this function:

$f(x) = \begin{cases} 0, & x \lt 1 \\ 1, & x = 1 \\ 0, & x > 1 \end{cases}$

According to Spivak's Calculus I, a function is integrable on $[a, b]$ (let's take $[0, 2]$ here) if $inf \{ U(f, P) \} = sup \{ U(f, P) \}$. Where $P$ is a partition of $[a, b]$ and $U$ means Upper areas, $L$ lower areas.

So, skipping some math, this means I have to show $U(f, P) - L(f, P) < ε$ for every $ε > 0$.

It actually easy to show and I don't need an explanation on the math, but my problem is that I can't imagine it. I mean, let's try to draw Upper and Lower areas of this function. First wi divide $[a, b]$ into a finite number of parts and then we draw rectangles who approximate the function. It is clear that Lowe areas are always zero, so also upper areas has to be zero. But this means the partition of $[a, b]$ should exclude the point $x = 1$, but if I exclude $x = 1$ then I'm not integrating over $[0, 2]$ but on $[0, 1)$ and $(1, 2]$.

Sorry for all the "buts"...

Can someone help me this. I really can't imagine it

 

[tnich]

Hi, I'm re-studying integrals and I got stuck with this problem. Actually the math beyond it is very clear but I still can figure it out.

Take this function:

$f(x) = \begin{cases} 0, & x \lt 1 \\ 1, & x = 1 \\ 0, & x > 1 \end{cases}$

According to Spivak's Calculus I, a function is integrable on $[a, b]$ (let's take $[0, 2]$ here) if $inf \{ U(f, P) \} = sup \{ U(f, P) \}$. Where $P$ is a partition of $[a, b]$ and $U$ means Upper areas, $L$ lower areas.

So, skipping some math, this means I have to show $U(f, P) - L(f, P) < ε$ for every $ε > 0$.

It actually easy to show and I don't need an explanation on the math, but my problem is that I can't imagine it. I mean, let's try to draw Upper and Lower areas of this function. First wi divide $[a, b]$ into a finite number of parts and then we draw rectangles who approximate the function. It is clear that Lowe areas are always zero, so also upper areas has to be zero. But this means the partition of $[a, b]$ should exclude the point $x = 1$, but if I exclude $x = 1$ then I'm not integrating over $[0, 2]$ but on $[0, 1)$ and $(1, 2]$.

Sorry for all the "buts"...

Can someone help me this. I really can't imagine it

What do you mean by an "upper area"?

 

[dRic2]

Sorry, I just found this pic online. The upper area is the yellow one, the lower is the green one (for a generic function).

 

[tnich]

Sorry, I just found this pic online. The upper area is the yellow one, the lower is the green one (for a generic function).

Suppose you choose an ε. Can you choose an interval containing x=1 such that the upper rectangle has an area less than ε?

 

[fresh_42]

I think, this is worth reading in this context:
https://www.physicsforums.com/insights/omissions-mathematics-education-gauge-integration/
especially w.r.t. the difference between Riemann and Lebesgue integrability.

 

[tnich]

Suppose you choose an ε. Can you choose an interval containing x=1 such that the upper rectangle has an area less than ε?

I think your problem might be in the definition of U(f, P). If you take U as the sum of the areas of the rectangles corresponding to partitition P, and then take inf U(f, P) to be the infimum over all partitions P, Spivak's definition of integrable makes sense.

 

[dRic2]

Suppose you choose an ε. Can you choose an interval containing x=1 such that the upper rectangle has an area less than ε?

Of curse you can.

Spivak's definition of integrable makes sense.

As I said, I understood it. I just can't imagine it. For example let it be $[0, 2] = [0, 1-\frac ε 3] + (1-\frac ε 3, 1+\frac ε 3] + (1+\frac ε 3, 2]$. It follows naturally that $U - L < ε$ for every $ε > 0$. But, since it has to be true for every ε, how can you draw it? You can't because you will end up drawing a vertical line in x = 1 (otherwise there will always be an infinitesimally small rectangle with area greater than zero) and if you draw a line in x = 1 than it means that $ε/3$ has to be 0... but ε can't be 0 because it has to be $ε > 0$

Maybe it a stupid thing but I can't understand it.

I think, this is worth reading in this context:
https://www.physicsforums.com/insights/omissions-mathematics-education-gauge-integration/
especially w.r.t. the difference between Riemann and Lebesgue integrability.

Thank you, I read it, but I'm not very familiar with non-Riemann integrals because I'm an engineering student. I recently decided to re-study calculus to get a deeper understanding, but I didn't even study Lebesgue integrals in my 3 years as undergrad... (and I don't think there will be any more mathematics classes)

 

[tnich]

I just can't imagine it. For example let it be $[0, 2] = [0, 1-\frac ε 3] + (1-\frac ε 3, 1+\frac ε 3] + (1+\frac ε 3, 2]$. It follows naturally that $U - L < ε$ for every $ε > 0$. But, since it has to be true for every ε, how can you draw it? You can't because you will end up drawing a vertical line in x = 1 (otherwise there will always be an infinitesimally small rectangle with area greater than zero) and if you draw a line in x = 1 than it means that $ε/3$ has to be 0... but ε can't be 0 because it has to be $ε > 0$

You seem to be imagining it quite vividly, and from your lucid explanation, I would guess you are quite good at math. What you seem to be stuck on is the concept of limits. This is the big leap that everyone has to make in understanding calculus. If you are comfortable with limits in other contexts (derivatives for example), you might try to draw an analogy with this problem. If not, keep working on those limit proofs until they start to make sense to you.

 

[dRic2]

Thank you.

What you seem to be stuck on is the concept of limits. This is the big leap that everyone has to make in understanding calculus.

I think you are right... I started to think I was missing something more profound and I thought of that too. I guess I will have to study limits again and more profoundly

 

[dRic2]

I think I solved my problem by seeing $\int^2_0 f(x) dx$ as

$\lim_{b \rightarrow 1^-} \int^b_0 f(x) dx + \int^1_1 f(x) dx + \lim_{a \rightarrow 1^+} \int^2_a f(x) dx = \lim_{b \rightarrow 1^-} \int^b_0 f(x) dx + 0 + \lim_{a \rightarrow 1^+} \int^2_a f(x) dx$

So the partition of $[0, 2]$ that satisfies $inf \{ U(f, P) \} = sup\{ L(f, P) \}$ is $P = [0, 1) + [1, 1] + (1, 2] = [0, 2]$

Is that correct?

 

[tnich]

I think I solved my problem by seeing $\int^2_0 f(x) dx$ as

$\lim_{b \rightarrow 1^-} \int^b_0 f(x) dx + \int^1_1 f(x) dx + \lim_{a \rightarrow 1^+} \int^2_a f(x) dx = \lim_{b \rightarrow 1^-} \int^b_0 f(x) dx + 0 + \lim_{a \rightarrow 1^+} \int^2_a f(x) dx$

So the partition of $[0, 2]$ that satisfies $inf \{ U(f, P) \} = sup\{ L(f, P) \}$ is $P = [0, 1) + [1, 1] + (1, 2] = [0, 2]$

Is that correct?

I think that works as a way of visualizing the limiting behavior. However, the definition of integrability that you cite really depends on partitions of segments with non-zero lengths. If you can make a zero length interval to deal with any discontinuity, then you can make the inf equal to the sup for any function.

 

[dRic2]

You are right. This seems to work even with a function with an infinite number of discontinuity... and that's wrong, of course. I'll keep thinking about it