# Thread: proving that an interval I = [a,b] has length |b-a|

1. ## proving that an interval I = [a,b] has length |b-a|

prove that an interval I = [a,b] has length |b-a|.

so this is an example in my book. So for the proof they use the indicator function on the interval I so that if the intervals we split up the interval I into are completely within the interval I then the value of the function at those points is 1 and if their are intervals where they are completely outside the interval I then the value of the function at those points is 0. Because of this the book argues that the difference between the upper and lower sums is at most twice the volume of a single cube (or the length of an interval in this case).

Un(I) - Ln(I) </= 2(1/2^N) where 1/2^N is the length of a side of a cube or in this case an interval. My book defines integrals using dyadic pavings with the dyadic cubes C(k,N) being the set of all x in R^n such that k_i/2^N </= x_i < (k_i + 1)/2^N for 1 </= i </= n. N is the level of the paving where larger N means smaller cubes.

now my book says that as N approaches infinite, the upper and lower sums converge to the same limit and therefore the indicator function over I is integrable and the interval I has a volume or length. My book leaves the computation of the length of I as an exercise.

i could not figure out how to compute this so i checked the student solutions manual whose solution i did not understand. they said that (1/2^N) ((2^N)|b-a| - 2) </= Ln(I) </= Volume (I) </= Un(I) </= (1/2^N) ((2^N)|b-a| +2) and when N approaches infinite both sides become just |b-a| proving that the length of the interval I is |b-a|.

what i'm confused about is how did they calculate (1/2^N) ((2^N)|b-a| - 2) and (1/2^N) ((2^N)|b-a| +2) in the first place and how did they know that they were smaller than Ln and large than Un respectively? It seems very arbitrary to me. It seems to me that i could stick in anything else instead of |b-a| let's say |b^3 - a^5| and letting N approach infinite i could "prove" that the length of I will be |b^3 - a^5|. i know this is definitely not the case so i am missing something here. my solution manual just presents those sets of inequalities above and does not explain at all how they arrived at them. if anyone could help explain how the book came up with that i would be very grateful!

2. Originally Posted by oblixps
prove that an interval I = [a,b] has length |b-a|.

so this is an example in my book. So for the proof they use the indicator function on the interval I so that if the intervals we split up the interval I into are completely within the interval I then the value of the function at those points is 1 and if their are intervals where they are completely outside the interval I then the value of the function at those points is 0. Because of this the book argues that the difference between the upper and lower sums is at most twice the volume of a single cube (or the length of an interval in this case).

Un(I) - Ln(I) </= 2(1/2^N) where 1/2^N is the length of a side of a cube or in this case an interval. My book defines integrals using dyadic pavings with the dyadic cubes C(k,N) being the set of all x in R^n such that k_i/2^N </= x_i < (k_i + 1)/2^N for 1 </= i </= n. N is the level of the paving where larger N means smaller cubes.

now my book says that as N approaches infinite, the upper and lower sums converge to the same limit and therefore the indicator function over I is integrable and the interval I has a volume or length. My book leaves the computation of the length of I as an exercise.

i could not figure out how to compute this so i checked the student solutions manual whose solution i did not understand. they said that (1/2^N) ((2^N)|b-a| - 2) </= Ln(I) </= Volume (I) </= Un(I) </= (1/2^N) ((2^N)|b-a| +2) and when N approaches infinite both sides become just |b-a| proving that the length of the interval I is |b-a|.

what i'm confused about is how did they calculate (1/2^N) ((2^N)|b-a| - 2) and (1/2^N) ((2^N)|b-a| +2) in the first place and how did they know that they were smaller than Ln and large than Un respectively? It seems very arbitrary to me. It seems to me that i could stick in anything else instead of |b-a| let's say |b^3 - a^5| and letting N approach infinite i could "prove" that the length of I will be |b^3 - a^5|. i know this is definitely not the case so i am missing something here. my solution manual just presents those sets of inequalities above and does not explain at all how they arrived at them. if anyone could help explain how the book came up with that i would be very grateful!
I'm still a little confused with what 'length' means here? Is this some kind of measure? Could you give a texed definition of 'length'?

3. oh sorry. the book defines the notion of n-dimensional volume but in this example we're talking about an interval so its just length. the definition my book gives is:
when 1_A (indicator function on set A) is integrable, the n-dimensional volume of A is vol A = integral [over R^n] of (1_A) |d^n x|.

4. Originally Posted by oblixps
oh sorry. the book defines the notion of n-dimensional volume but in this example we're talking about an interval so its just length. the definition my book gives is:
when 1_A (indicator function on set A) is integrable, the n-dimensional volume of A is vol A = integral [over R^n] of (1_A) |d^n x|.
And $|d^n x|$ is?

5. that's the author's way of denoting dV or dx1 dx2 dx3... dxn. the absolute value bars are used by the author to mean that orientation of the domain does not matter. this is only the introductory lesson in the integration chapter so he has not even defined what orientation of the domain is yet.

6. So, I suppose you would be looking at the length of an interval as a 1-dimensional volume?

In that case, using the Riemann integral (don't really know what else to use...), the answer should be immediate, should it not?

7. well im trying to use the definition of upper and lower sums since my book did not even cover techniques of integration yet. Ln = sum of [(volume of a cube)(inf (f) on each cube)] and Un = sum of [(volume of a cube)(sup(f) on each cube)]. don't know if this is right or not but i got Un to be sum of 2^N / 2^N and Ln to be (2^N-2)/2^N. but im still not sure how they book got (1/2^N) ((2^N)|b-a| - 2) </= Ln(I) from.

8. Basically you're integrating the indicator function, which is identically 1 on your interval $[a,b]$. So on any "cube" (subinterval) contained in your interval, the infimum and supremum of your function should both be exactly 1.

Not really sure where the $2^n$ are coming from though...

9. oh ok so the inf and sup of the function are both 1 on the interval. so the lower sum would be (1)(volume of one cube) and the volume of one cube would be 1/2^N according to my book. but then both the lower and upper sum would be 1/2^N. then (1/2^N) ((2^N)|b-a| - 2) would be less than the lower sum which is 1/2^N but how did the book come up with ((2^N)|b-a| - 2) and how did they know that ((2^N)|b-a| - 2) is less than 1? sorry for asking so many questions.