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

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

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


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.


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.

  • 4
    $\begingroup$ This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $\implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does. $\endgroup$ – Aloizio Macedo Aug 13 at 21:10
  • $\begingroup$ Good point. Replace "Lebesgue" with "Riemann" as done in your answer. $\endgroup$ – Hans Engler Aug 14 at 13:31

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