# Riemann Integrable function

(From I.S.I. M.Math Subjective Sample Paper 2013)

Let N>0 and let $$\mathbf{ f:[0,1] to [0,1] }$$ be denoted by f(x) = 1 if x=1/i for some integer $$\mathbf{i\le N}$$ and f(x) = 0 for all other values of x. Show that f is Riemann Integrable.

Discussion

First let’s get the notations in place (Riemann integral has several notations in different books).
Let P be a tagged partition of [0,1] that is $$\mathbf{wp = {([x_{i-1} , x_i ], t_i)}_{i=1}^n }$$.

S(f,P) be the Riemann Sum of function f given this tagged partition; that is  $$\mathbf{ S(f, wp) = \sum_{i=1}^n f(t_i)(x_i -x_{i-1}) }$$

We conjecture that the Riemann Integral of the given function is 0 (how do we know it? A guess. If we wish to eliminate this guessing step, then we have to use Cauchy criterion for the proof).

We show that $$\mathbf{ S(f, wp) < \epsilon}$$ for any $$\mathbf{ \epsilon > 0 }$$ (that is we will be able to find a $$\mathbf{\delta_{\epsilon}}$$ which is the norm of a partition concerned)

Let us take $$\mathbf{\delta_{\epsilon} = \frac{\epsilon}{2N} }$$ that is we divide [0,1] into $$\mathbf{\lfloor \frac{2N}{\epsilon} \rfloor }$$ parts of equal length. The Riemann sum of the given function over this partition is at most $$\mathbf{\frac{\epsilon}{2} }$$ which is smaller than $latex \mathbf{\epsilon}$

Proved