**(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**

## No comments, be the first one to comment !