Regarding Riemann integration defination

[seshikanth]

Regarding "Riemann integration defination"

Hi,

I did not understand the following:

We have : Partition is always a "finite set".
A function f is said to Riemann integrable if f is bounded and
Limit _{||P|| -> 0} L(f,P) = Limit_{||P|| -> 0} U(f,P)
where L(f,P) and U(f,P) are lower and upper Darbaux Sums, ||P|| is the norm of Partition.

If ||P|| -> 0 then we have max sub-interval length = 0 => we have infinite points in Partition which leads to contradiction of definition of Partition being "FINITE SET".
I think i am missing something here!

Thanks,

 

[lavinia]

Hi,

I did not understand the following:

We have : Partition is always a "finite set".
A function f is said to Riemann integrable if f is bounded and
Limit _{||P|| -> 0} L(f,P) = Limit_{||P|| -> 0} U(f,P)
where L(f,P) and U(f,P) are lower and upper Darbaux Sums, ||P|| is the norm of Partition.

If ||P|| -> 0 then we have max sub-interval length = 0 => we have infinite points in Partition which leads to contradiction of definition of Partition being "FINITE SET".
I think i am missing something here!

Thanks,

the number of points in the partition increases without bound but is never infinite.

There is no limiting partition, only a limiting value for the Riemann sums.

 

[seshikanth]

Can you please elaborate? The norm of partition tending to zero implies that the number of points in the partition also tends to infinite.

Thanks,

 

[lavinia]

Can you please elaborate? The norm of partition tending to zero implies that the number of points in the partition also tends to infinite.

Thanks,

yes but each partition is finite. As the norm of the partition shrinks the number of finite points in it increases without bound - but it is never infinite. In the limit there is no partition.

Example: integrate the function f(x) = x over the unit interval. A typical Riemann sum with n partition points looks like

(1 + ... + n-1)/n^2 = 1/2(n-1)n/n^2 = 1/2(n-1)/n = 1/2( 1-1/n). As n goes to infinity this goes to 1/2.

 

[seshikanth]

If the norm shrinks and shrinks tending towards zero, How can the number of points in the partition be finite? This is where i am thinking! (Lower Darbaux sum will be equal to Upper Darbaux sum only when the number of intervals tends to infinite)

 

[lavinia]

If the norm shrinks and shrinks tending towards zero, How can the number of points in the partition be finite? This is where i am thinking! (Lower Darbaux sum will be equal to Upper Darbaux sum only when the number of intervals tends to infinite)

I just added an example.

The numbers 1,2 ...n, n+1, ... tend to infinity but never get there. That is the same for the number of partition points

 

[seshikanth]

Got it!

 

[seshikanth]

One more question: Please correct me if i am wrong here-
"If a function f is Riemann integrable on closed and bounded interval [a,b] then f is continuous on [a,b] and also the converse is also true"

i.e., Riemann integrable <=> f being continuous

 

[lavinia]

One more question: Please correct me if i am wrong here-
"If a function f is Riemann integrable on closed and bounded interval [a,b] then f is continuous on [a,b] and also the converse is also true"

i.e., Riemann integrable <=> f being continuous

no. It does not have to be continuous. But the set of discontinuities must have measure zero.

Try calculating the Riemann sums for the function that is 1 from 0 to 1/2 and 2 from 1/2 to 1.

Intuitively you can see that the area under this discontinuous function is the area under two rectangles.