Can it be possible?

Calculus Level 3

Does there exist a function $f: [ -1, 1 ] \rightarrow \mathbb{R}$ that is not Riemann integrable, but $f^2$ (defined as $f^2(x) = (f(x))^2$ for all $x$ ) is Riemann integrable?

Yes, there exists No, there doesn't

1 solution

Guillermo Templado
Oct 13, 2015

Let's consider f:[-1,1]------->R to be defined as f(x)= 1 if x is a rational number in [-1,1] and f(x)=-1 if x is an irrational number in [-1,1]. This function is no-integrable Riemann because the set of discontinuity points of f doesn't have zero measure, however, (f^2)(x)=1 if x is in [-1,1] and this function is Riemann integrable

This is @Calvin Lin 's favorite non-integrable function

Agnishom Chattopadhyay - 5 years, 8 months ago

Any discontinuous function (whose set of discontinuities is not of zero Lebesgue measure) that has a range $\{ -a, a \}$ , $a \neq 0$ will satisfy the conditions.

Pranshu Gaba - 5 years, 8 months ago

Can it be possible?

1 solution

0 pending reports