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*}<br /> M_k (x_k-x_{k-1}) - m_k (x_k-x_{k-1}) \\<br /> &amp;=(M_k - m_k)(x_k-x_{k-1}) \\<br /> &amp;=(x_k^2-x_{k-1}^2)(x_k-x_{k-1}) \\ <br /> &amp;=(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&lt;x_1&lt;...&lt;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&lt;x_1=a+\frac{d}{2^i}&lt;x_1=a+2 \frac{d}{2^i}&lt;...&lt;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*}<br /> \left( U(f,P_k)-L(f,P_k) \right) \\<br /> &amp;=\sum_{j=1}^n (M_j-m_{j-1})(x_j-x_{j-1}) \\<br /> &amp;=\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}<br />.

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&lt;x_1=a+\frac{d}{2^i}&lt;x_1=a+2 \frac{d}{2^i}&lt;...&lt;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}) &lt;\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&lt;x_1&lt;x_2&lt;...&lt;x_{n-1}&lt;x_n}
The x_ks 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&lt;x_1=a+\frac{b-a}{k}&lt;...&lt;x_j=a+j\frac{b-a}{k}&lt;...&lt;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*}<br /> lim \left(U(f)-L(f) \right)&amp;=lim\left( U(f,P_k)-L(f,P_k) \right) \\<br /> &amp;=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}) \\<br /> &amp;=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) \\<br /> &amp;=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\<br /> &amp;=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) \\<br /> &amp;=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\<br /> \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*}&amp;= 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) \\<br /> &amp;= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\<br /> &amp;= 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) \\<br /> &amp;=lim_{k \rightarrow \infty } ((k^2+k) +2a) \frac{(b-a)^3}{k^3} \\<br /> &amp;=lim_{k \rightarrow \infty } (k^2 +2a) \frac{(b-a)^3}{k^3}\\<br /> &amp;=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} + \frac{2a(b-a)^3}{k^3}\\<br /> &amp;= 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.