A function f that is not Riemann integrable but |f| is Riemann integrable?

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A function can be constructed that is not Riemann integrable while its absolute value is Riemann integrable. The example discussed is f(x) = 1 for rational x and f(x) = -1 for irrational x on the interval [0,1]. This function is not Riemann integrable due to its discontinuities, as it oscillates between two values without settling. However, the absolute value |f| equals 1 for all x, which is constant and thus Riemann integrable. This illustrates the distinction between the integrability of a function and its absolute value.
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I was just going over Riemann integrability and how to prove it, and was just wondering is it possible to have a function f that is not Riemann integrable but |f| is Riemann integrable? Say on an interval [0,1] for example. (as that is what most examples I have done are on so easiest for me to compare)
How would this work? Or does it not?
 
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f(x)=1 for rational x
f(x)=-1 for irrational x

Is it Riemann integrable?
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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