# Clarification on Integrability

by all.over   Last Updated August 13, 2019 21:20 PM

If $$f$$ in integrable on some interval $$[a,b]$$ then we know that $$\lvert f \rvert$$ is also integrable on that same interval.

There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $$f$$ where

$$\displaystyle \int_0^1 f dx = \lim_{c \downarrow 0} \int_c^1 fdx$$

exists and yet for $$\lvert f \rvert$$ this limit fails to exist.

How does this not contradict the implication above?

One such constuction is to set $$f(x) = (-1)^{k+1}(k+1), \forall x \in (\frac{1}{k+1},\frac{1}{k}]$$.

Tags :

The implication $$f$$ is integrable $$\Rightarrow$$ $$|f|$$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.

Hans Engler
August 13, 2019 20:34 PM

There is no contradiction because $$f$$ is not Riemann-integrable on $$[0,1]$$. The fact that the limit $$\lim_{c \to 0}\int_c ^1f(x)dx$$ exists does not mean that $$f$$ is Riemann-integrable on $$[0,1]$$. Note that Rudin says in the exercise that we can define the symbol $$\int_0^1f(x)dx$$ to mean said limit in the case where we have a function on $$(0,1]$$. It does not mean that $$f$$ is Riemann-integrable on $$[0,1]$$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.

PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $$\int_ 0^1f(x)dx$$ agree when $$f$$ is Riemann-integrable.

Aloizio Macedo
August 13, 2019 21:02 PM

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