Given a function and a closed interval such that
for all .
- Prove is continuous at every point .
- If is integrable on prove
- If is any point prove
- Proof. Let be any point in . Then, for any let . Then
Thus, for any , we have whenever . Hence,
Therefore, is continuous at every point
- Proof. Since is a constant (the value of the function evaluated at the constant ), we have
Therefore, we can compute,
The final inequality follows since, using the monotone property of the integral,
Then, since by our hypothesis on , we have
- Proof. The proof proceeds similarly to that of part (b),
Then, since , we break the integral into two pieces (to deal with the fact that for and for ):
But now, we know implies , so we have the inequality,