Is the function riemann integrable? havent got a clue

[natasha d]

Homework Statement

f:[0,1]→R where f(x)= 0 if x=0 and f(x)=1/n when 1/(1+n) < x ≤ 1/n, n \in N.
is f Riemann integrable

Homework Equations

R integrable only when L(f) =U(f)
L(f) = largest element of the set of lower sums for n partitions
U(f) = least element of the set of upper sums for n partitions

f is R integrable if the function has finite discontinuities

The Attempt at a Solution

tried plotting a graph, where the function is constant on intervals [1/(1+n) , 1/n] since all x in between take the value 1/n
that gives me intervals on which f is R integrable except at 1/(1+n)
but that also gives ∞ such R integrable intervals, which means ∞ discontinuities...
On an interval the lower sum = the upper sum
= length of the interval X F(x)
= [(1/n)-(1/(1+n))] X (1/n)

the total integral will be the summation with n → ∞ ?

 

[chiro]

Hey natasha d and welcome to the forums.

It looks like this function for each realization of n (i.e. each value of the natural numbers) will simply be a box in the given interval.

You say that it's defined to be 0 at x = 0, but what about outside the interval of the box (i.e. outside of (1/(n+1),1/n]? Because n will be constant for a particular realization of a natural number, it means the interval will be finite which means that it will look more or less like a single box with two discontinuities if it is defined everywhere else to be zero.

The question I have is, "Is the function defined to be zero everywhere outside the above interval, or does it have a different definition?"

 

[natasha d]

what 'box'? um.. on the interval [0,1] f(x)=0 only at x=0, everywhere else the value of f for an x will be the 1/n value x is less than or equal to..
i thought that meant in, say, [1/2,1], f(x)=1 , except at x=1/2
did i miss something?

 

[HallsofIvy]

There is a fundamental theorem that says that a function is Riemann integrable on an interval if and only if it has only "jump" discontinuities and the set of discontinuities on the interval has (Lesbesque measure) 0. Here the set of discontinuities is the set of numbers 1/n which is countable and so has measure 0.