#### To determine

**To show:**

f is not integrable on 0, 1

**Proof:**

#### Explanation

**1) Concept:**

i. If f is integrable, then thefollowing limit must exist

limn→∞∑i=1nfxi* ∆x

Where,

∆x=b-an when xi* is in [a+(i-1) ∆x,a+i ∆x]

ii. Theorem:If f is continuous on a, b, or if f has only a finite number of jump discontinuities, then f is integrable on a, b

2) **Calculation:**

We have an interval 0, 1,

So divide it into n intervals, such that ∆x=b-an=1n

The interval in,i+1n always contains a rational and irrational number, however large n is

Let us take xi* as the irrational number then

limn→∞∑i=1nfxi* ∆x

=limn→∞∑i=1n1·1n

=limn→∞1n∑i=1n 1

=limn→∞1n·n

=limn→∞1

=1

Now, take xi* as the rational number then

limn→∞∑i=1nfxi* ∆x

=limn→∞∑i=1n0·1n

=limn→∞1n∑i=1n0

=limn→∞1n·0

=limn→∞0

=0

So here, thelimit depends on thevalue of xi*

Thus the limit which gives integral doesn’t exist. Hence the integral doesn’t exists, that is the function is not integrable.Another proof is: Notice that this function has jump discontinuity at every rational number and irrational number. That is it has infinitely many jump discontinuities. So it is not integrable.

**Conclusion:**

f is not integrable on 0, 1