- #1

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i just wanted to ask how would one prove that a function is riemann integrable through the definition that the lower integral has to equal the upper integral, an example on the function f(x)=x^2 would be of great help

thanxs

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- #1

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i just wanted to ask how would one prove that a function is riemann integrable through the definition that the lower integral has to equal the upper integral, an example on the function f(x)=x^2 would be of great help

thanxs

- #2

Hurkyl

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Well, start off by writing exactly what you want to prove. (In formulas, I mean, not words)

- #3

mathwonk

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this follows from the formula for the sum of the first n squares of integers.

- #4

rachmaninoff

Here's a rough outline of what you would need to work through:

Observe that [tex]f(x)=x^2[/tex] is strictly increasing on [tex] \left[ 0,\infty) [/tex], so for any subinterval [tex] \left[ x_{k-1},x_k \right) \epsilon \left[ 0,\infty \right) [/tex], you have

[tex] m_k = f(x_{k-1}) = x_{k-1}^2[/tex]

[tex] M_k = f(x_{k}) = x_{k}^2[/tex]

[tex] \begin{align*}

M_k (x_k-x_{k-1}) - m_k (x_k-x_{k-1}) \\

&=(M_k - m_k)(x_k-x_{k-1}) \\

&=(x_k^2-x_{k-1}^2)(x_k-x_{k-1}) \\

&=(2 x_{k-1}(x_k-x_{k-1}) + (x_k-x_{k-1})^2)(x_k-x_{k-1}) \end{align}[/tex]

where we are partioning an interval [a,b] with an arbitrary partition [tex]P=\{a=x_0<x_1<...<x_n=b\}[/tex]. To show that f(x) is Reimann integrable, you must show that the lower integral is equal to the upper integral, where

*infimum* instead of a *supremum*). An easy way to get the inf/sup is to define your own sequence of partitions [tex](P_k)[/tex]: if it converges, then you can replace the inf/sup with the limit of the sequence of

Hint:

[tex] \begin{align*}

\left( U(f,P_k)-L(f,P_k) \right) \\

&=\sum_{j=1}^n (M_j-m_{j-1})(x_j-x_{j-1}) \\

&=\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \end{align}

[/tex].

You'll need to modify this proof slightly for the x<0 case (where is this assumption made?) and then figure out how to combine the results to show that [tex]f(x)=x^2[/tex] is Reimann-integrable over all of**R**.

Observe that [tex]f(x)=x^2[/tex] is strictly increasing on [tex] \left[ 0,\infty) [/tex], so for any subinterval [tex] \left[ x_{k-1},x_k \right) \epsilon \left[ 0,\infty \right) [/tex], you have

[tex] m_k = f(x_{k-1}) = x_{k-1}^2[/tex]

[tex] M_k = f(x_{k}) = x_{k}^2[/tex]

[tex] \begin{align*}

M_k (x_k-x_{k-1}) - m_k (x_k-x_{k-1}) \\

&=(M_k - m_k)(x_k-x_{k-1}) \\

&=(x_k^2-x_{k-1}^2)(x_k-x_{k-1}) \\

&=(2 x_{k-1}(x_k-x_{k-1}) + (x_k-x_{k-1})^2)(x_k-x_{k-1}) \end{align}[/tex]

where we are partioning an interval [a,b] with an arbitrary partition [tex]P=\{a=x_0<x_1<...<x_n=b\}[/tex]. To show that f(x) is Reimann integrable, you must show that the lower integral is equal to the upper integral, where

[tex]L(f)=sup \{ \sum_{k=1}^n m_k(x_k-x_{k-1}) [/tex]

over all partitions P, and U(f) is defined parallel-y (as an [tex] \sum_{k=1}^n m_k(x_k-x_{k-1}) [/tex]

defined over the [tex](P_k)[/tex]. For example try defining [tex]d=b-a[/tex] and [tex]P_i=\{a=x_0<x_1=a+\frac{d}{2^i}<x_1=a+2 \frac{d}{2^i}<...<x_{2^i}=b\}[/tex]. So in the k-th partition, [tex]x_i= i \frac{d}{2^i}[/tex]. Then figure out what the terms of [tex] | \left( L(f,P_k)-U(f,P_k) \right) | [/tex] look like, and determine if they converge; if they do, then you can show that [tex]U(f)-L(f)=0[/tex] (can you see how?)Hint:

[tex] \begin{align*}

\left( U(f,P_k)-L(f,P_k) \right) \\

&=\sum_{j=1}^n (M_j-m_{j-1})(x_j-x_{j-1}) \\

&=\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \end{align}

[/tex].

You'll need to modify this proof slightly for the x<0 case (where is this assumption made?) and then figure out how to combine the results to show that [tex]f(x)=x^2[/tex] is Reimann-integrable over all of

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- #5

mathwonk

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is it that hard? for x^2 on [0,a] you have a subdivision of length a/n, and the upper value on [(k-1)a/n, k/n] is (ka)^2/n^2, while the lower value is [(k-1)a]^2/n^2.

so the lower sum is (a/n)[0^2/n^2 + a^2/n^2 +.......+[(n-1)a]^2/n^2],

and the upper sum is (a/n)[a^2/n^2 + .....+[(n-1)a]^2/n^2 + [(n)a]^2/n^2].

thus the difference between upper and lower sums is (a/n)[(n)a]^2/n^2 = a^3/n, which goes to zero as n goes to infinity. done.

i'm a little sleepy but isn't this it?

by the way this shows that what i said before is not needed. i.e. you only need that formula to evaluate the integral. existence is much easier.

so the lower sum is (a/n)[0^2/n^2 + a^2/n^2 +.......+[(n-1)a]^2/n^2],

and the upper sum is (a/n)[a^2/n^2 + .....+[(n-1)a]^2/n^2 + [(n)a]^2/n^2].

thus the difference between upper and lower sums is (a/n)[(n)a]^2/n^2 = a^3/n, which goes to zero as n goes to infinity. done.

i'm a little sleepy but isn't this it?

by the way this shows that what i said before is not needed. i.e. you only need that formula to evaluate the integral. existence is much easier.

Last edited:

- #6

rachmaninoff

Well, yes. I just threw in a bit of extra rigor - our end results are identical.i'm a little sleepy but isn't this it?

- #7

mathwonk

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oh is ee,l you gave a lot of detail i assumed as obvious.

- #8

rachmaninoff

Your proof is plenty rigorous, no offence intended.

- #9

mathwonk

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you are right. i am not as energetic as you are tonight. good job. peace

- #10

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so far i have worked my way through to get thisrachmaninoff said:An easy way to get the inf/sup is to define your own sequence of partitions [tex](P_k)[/tex]: if it converges, then you can replace the inf/sup with the limit of the sequence of

[tex] \sum_{k=1}^n m_k(x_k-x_{k-1}) [/tex]defined over the [tex](P_k)[/tex]. For example try defining [tex]d=b-a[/tex] and [tex]P_i=\{a=x_0<x_1=a+\frac{d}{2^i}<x_1=a+2 \frac{d}{2^i}<...<x_{2^i}=b\}[/tex]. So in the k-th partition, [tex]x_i= i \frac{d}{2^i}[/tex]. Then figure out what the terms of [tex] | \left( L(f,P_k)-U(f,P_k) \right) | [/tex] look like, and determine if they converge.

[tex]U(f)-L(f) \le U(f,P_k) - L(f,P_k) = \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) [/tex]

but i dont understand how how you could make up a sequence of partitions, to get this

[tex]\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) <\epsilon[/tex]

well especially the the sequence of partition you used i dont understand it, like what is a sequence of partitions isnt it just an ordered set of partitions?

please help

thanxs

- #11

rachmaninoff

Consider the set of all partitions {P} of [a,b], defined as

[tex]{P:P=\{a=x_0<x_1<x_2<...<x_{n-1}<x_n}[/tex]

The [tex]x_k[/tex]s divide your interval into a countable number of sub-intervals [tex] \left[ x_{k-1}, x_k \right] [/tex]. The [tex]definition[/tex] of the lower/upper integral is rather abstract and useless - it defines them as the supremum/infimum of [tex]L(f,P)[/tex] and [tex]U(f,P)[/tex] respectively on the set of*all possible* partitions P. But there exists a nice lemma that if you take a sequence of partitions [tex](P_k)[/tex], and the limit of their partition sizes [tex]\delta = x_{k}-x_{k-1}[/tex] goes to zero (I think that's the requirement, it might be more general), then the supremum (infimum) of the [tex]L(f,P)[/tex] or [tex]U(f,P)[/tex] is the same as the limit of the sequence [tex]U(f,p_k)[/tex]:

sup{[tex]L(f,P):P[/tex] is a partition of [a,b]}=lim [tex] U(f,p_k)[/tex]

Intuitively, as your partition gets "finer and finer", the error of the lower (upper) sum gets less. I forgot how the rigorous version goes here. The point is, instead of working with supremums, you work with a limits of sequences (which are 'nicer').

is unique to f(x)=x^2, under any partition (I derived at the top of my first post, assuming x>0 and that x^2 is strictly increasing; you get something very similiar when x<0).

To actually work with this, it would be nice to actually choose a sequence of partitions (that gets succesively finer): an easy one is

[tex]P_k=P \{ a=x_0<x_1=a+\frac{b-a}{k}<...<x_j=a+j\frac{b-a}{k}<...<x_k=a+k\frac{b-a}{k}=b \} [/tex]

Which reduces down to what**mathwonk** was using: (I'll quote his post)

[tex]\begin{align*}

lim \left(U(f)-L(f) \right)&=lim\left( U(f,P_k)-L(f,P_k) \right) \\

&=lim \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \\

&=lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\

&=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\

&=0 \end{align}[/tex]

[tex]{P:P=\{a=x_0<x_1<x_2<...<x_{n-1}<x_n}[/tex]

The [tex]x_k[/tex]s divide your interval into a countable number of sub-intervals [tex] \left[ x_{k-1}, x_k \right] [/tex]. The [tex]definition[/tex] of the lower/upper integral is rather abstract and useless - it defines them as the supremum/infimum of [tex]L(f,P)[/tex] and [tex]U(f,P)[/tex] respectively on the set of

sup{[tex]L(f,P):P[/tex] is a partition of [a,b]}=lim [tex] U(f,p_k)[/tex]

Intuitively, as your partition gets "finer and finer", the error of the lower (upper) sum gets less. I forgot how the rigorous version goes here. The point is, instead of working with supremums, you work with a limits of sequences (which are 'nicer').

I'm not entirely sure what you're asking about - let's see, the [tex]\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) [/tex]but i dont understand how how you could make up a sequence of partitions, to get this

[tex]U(f)-L(f) \le U(f,P_k) - L(f,P_k) = \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) [/tex]

is unique to f(x)=x^2, under any partition (I derived at the top of my first post, assuming x>0 and that x^2 is strictly increasing; you get something very similiar when x<0).

To actually work with this, it would be nice to actually choose a sequence of partitions (that gets succesively finer): an easy one is

[tex]P_k=P \{ a=x_0<x_1=a+\frac{b-a}{k}<...<x_j=a+j\frac{b-a}{k}<...<x_k=a+k\frac{b-a}{k}=b \} [/tex]

Which reduces down to what

i.e.,for x^2 on [0,a] you have a subdivision of length a/n, and the upper value on [(k-1)a/n, k/n] is (ka)^2/n^2, while the lower value is [(k-1)a]^2/n^2.

so the lower sum is (a/n)[0^2/n^2 + a^2/n^2 +.......+[(n-1)a]^2/n^2],

and the upper sum is (a/n)[a^2/n^2 + .....+[(n-1)a]^2/n^2 + [(n)a]^2/n^2

[tex]\begin{align*}

lim \left(U(f)-L(f) \right)&=lim\left( U(f,P_k)-L(f,P_k) \right) \\

&=lim \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \\

&=lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\

&=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\

&=0 \end{align}[/tex]

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- #12

rachmaninoff

Hope this is making some sense...

- #13

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I have to admit this sequence stuff certainly makes things look nice, or else I would have dwelled within the supremum and infimum stuff, well last of all i cant see how you made this step?

rachmaninoff said:i.e.,

[tex]lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\

&=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\

\end{align}[/tex]

- #14

rachmaninoff

From the definition of the particular partition sequence [tex](P_k)[/tex] I'm using,

[tex]x_j=a+j\frac{b-a}{k}[/tex]

and so

[tex]\begin{align*}&= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\

&= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\

&= lim \sum_{j=1}^{k}\left( 2j\frac{b-a}{k} +2a\left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\

&=lim_{k \rightarrow \infty } ((k^2+k) +2a) \frac{(b-a)^3}{k^3} \\

&=lim_{k \rightarrow \infty } (k^2 +2a) \frac{(b-a)^3}{k^3}\\

&=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} + \frac{2a(b-a)^3}{k^3}\\

&= 0 \end{align}[/tex]

Apparently I messed up slightly - I copied mathwonk's result, and his interval was [0,b] and mine was [a,b], hence the extra term in the expressions - anyway it all goes to zero.

edit: the "formula for the sum of the first n squares of integers" is how you get rid of the sum in the 5th line.

[tex]x_j=a+j\frac{b-a}{k}[/tex]

and so

[tex]\begin{align*}&= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\

&= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\

&= lim \sum_{j=1}^{k}\left( 2j\frac{b-a}{k} +2a\left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\

&=lim_{k \rightarrow \infty } ((k^2+k) +2a) \frac{(b-a)^3}{k^3} \\

&=lim_{k \rightarrow \infty } (k^2 +2a) \frac{(b-a)^3}{k^3}\\

&=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} + \frac{2a(b-a)^3}{k^3}\\

&= 0 \end{align}[/tex]

Apparently I messed up slightly - I copied mathwonk's result, and his interval was [0,b] and mine was [a,b], hence the extra term in the expressions - anyway it all goes to zero.

edit: the "formula for the sum of the first n squares of integers" is how you get rid of the sum in the 5th line.

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