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 $f(x) = (-1)^{k+1}(k+1), \forall x \in (\frac{1}{k+1},\frac{1}{k}]$.

Answers 2

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
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
Aloizio Macedo
August 13, 2019 21:02 PM

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