Proving Riemann Integrability of f(x)=x^2

[steven187]

hello all

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

 

[Hurkyl]

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

 

[mathwonk]

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

 

[rachmaninoff]

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

Observe that $$f(x)=x^2$$ is strictly increasing on $$\left[ 0,\infty)$$, so for any subinterval $$\left[ x_{k-1},x_k \right) \epsilon \left[ 0,\infty \right)$$, you have
$$m_k = f(x_{k-1}) = x_{k-1}^2$$
$$M_k = f(x_{k}) = x_{k}^2$$
$$\begin{align*} M_k (x_k-x_{k-1}) - m_k (x_k-x_{k-1}) \\ &=(M_k - m_k)(x_k-x_{k-1}) \\ &=(x_k^2-x_{k-1}^2)(x_k-x_{k-1}) \\ &=(2 x_{k-1}(x_k-x_{k-1}) + (x_k-x_{k-1})^2)(x_k-x_{k-1}) \end{align}$$
where we are partioning an interval [a,b] with an arbitrary partition $$P=\{a=x_0<x_1<...<x_n=b\}$$. To show that f(x) is Reimann integrable, you must show that the lower integral is equal to the upper integral, where

$$L(f)=sup \{ \sum_{k=1}^n m_k(x_k-x_{k-1})$$​

over all partitions P, and U(f) is defined parallel-y (as an infimum instead of a supremum). An easy way to get the inf/sup is to define your own sequence of partitions $$(P_k)$$: if it converges, then you can replace the inf/sup with the limit of the sequence of

$$\sum_{k=1}^n m_k(x_k-x_{k-1})$$​

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

Hint:
$$\begin{align*} \left( U(f,P_k)-L(f,P_k) \right) \\ &=\sum_{j=1}^n (M_j-m_{j-1})(x_j-x_{j-1}) \\ &=\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \end{align}$$.

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 $$f(x)=x^2$$ is Reimann-integrable over all of R.

 

[mathwonk]

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.

 

[rachmaninoff]

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

Well, yes. I just threw in a bit of extra rigor - our end results are identical.

 

[mathwonk]

extra rigor? where is my argument unrigorous?

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

 

[rachmaninoff]

Well, the OP asked for proof 'from defintion', so I thought to throw in , you know, how to go from a superemum over partitions to a limit of any convergent sequence of partitions, and then to choose a nice-looking sequence - it's stuff that wasn't obvious to me when I took first-year real analysis (lots of lemmas were involved, iirc).

Your proof is plenty rigorous, no offence intended.

 

[mathwonk]

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

 

[steven187]

An easy way to get the inf/sup is to define your own sequence of partitions $$(P_k)$$: if it converges, then you can replace the inf/sup with the limit of the sequence of

$$\sum_{k=1}^n m_k(x_k-x_{k-1})$$​

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

so far i have worked my way through to get this
$$U(f)-L(f) \le U(f,P_k) - L(f,P_k) = \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1})$$
but i don't understand how how you could make up a sequence of partitions, to get this
$$\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) <\epsilon$$
well especially the the sequence of partition you used i don't understand it, like what is a sequence of partitions isn't it just an ordered set of partitions?

please help

thanxs

 

[rachmaninoff]

Consider the set of all partitions {P} of [a,b], defined as
$${P:P=\{a=x_0<x_1<x_2<...<x_{n-1}<x_n}$$
The $$x_k$$s divide your interval into a countable number of sub-intervals $$\left[ x_{k-1}, x_k \right]$$. The $$definition$$ of the lower/upper integral is rather abstract and useless - it defines them as the supremum/infimum of $$L(f,P)$$ and $$U(f,P)$$ respectively on the set of all possible partitions P. But there exists a nice lemma that if you take a sequence of partitions $$(P_k)$$, and the limit of their partition sizes $$\delta = x_{k}-x_{k-1}$$ goes to zero (I think that's the requirement, it might be more general), then the supremum (infimum) of the $$L(f,P)$$ or $$U(f,P)$$ is the same as the limit of the sequence $$U(f,p_k)$$:
sup{$$L(f,P):P$$ is a partition of [a,b]}=lim $$U(f,p_k)$$
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').

but i don't understand how how you could make up a sequence of partitions, to get this
$$U(f)-L(f) \le U(f,P_k) - L(f,P_k) = \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1})$$

I'm not entirely sure what you're asking about - let's see, the $$\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1})$$
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 similar 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
$$P_k=P \{ a=x_0<x_1=a+\frac{b-a}{k}<...<x_j=a+j\frac{b-a}{k}<...<x_k=a+k\frac{b-a}{k}=b \}$$
Which reduces down to what mathwonk was using: (I'll quote his post)

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

i.e.,
$$\begin{align*} lim \left(U(f)-L(f) \right)&=lim\left( U(f,P_k)-L(f,P_k) \right) \\ &=lim \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \\ &=lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\ &=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\ &=0 \end{align}$$

 

[rachmaninoff]

Hope this is making some sense...

 

[steven187]

hello there

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 can't see how you made this step?

i.e.,
$$lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\ &=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\ \end{align}$$

 

[rachmaninoff]

From the definition of the particular partition sequence $$(P_k)$$ I'm using,
$$x_j=a+j\frac{b-a}{k}$$
and so
$$\begin{align*}&= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\ &= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\ &= lim \sum_{j=1}^{k}\left( 2j\frac{b-a}{k} +2a\left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\ &=lim_{k \rightarrow \infty } ((k^2+k) +2a) \frac{(b-a)^3}{k^3} \\ &=lim_{k \rightarrow \infty } (k^2 +2a) \frac{(b-a)^3}{k^3}\\ &=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} + \frac{2a(b-a)^3}{k^3}\\ &= 0 \end{align}$$

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.