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