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This was written in Freitag's (p196)

A continuous function $f:[a,b) \rightarrow \mathbb{C}$ is called improperly integrable iff the limit $$ \lim_{t\rightarrow b} \int_a^t f(x) \, dx $$ exists. It is (improperly) absolutely integralbe, if the above limit exists with $f$ replaced by $|f|$. Absolute integrability implies integrability (?)

What I don't understand is why the last holds bolded line holds.


My thoughts: I don't even see how this holds if $f$ were real. If we define $G(t):= \int_a^t f(x) \, dx$, then $G(t)$ is a continuous (by FTC), and bounded (by absolutely integrability) function on $[a,b)$. We need to show $\lim_{t \rightarrow b} G(t)$ exists, but it can be the case that $G(t)$ is some function like $\sin (1/x)$.

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Outline: imitate the proof for absolute convergence $\implies$ convergence for sums. Write down the Cauchy criterion for $g(x) = \int_0^x \lvert f \rvert$, and use it and the triangle inequality to show that $h(x) = \int_0^x f$ is Cauchy.

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  • $\begingroup$ Thanks a lot. Now I have a new question, does the Improper Integral (if exists) coincide with the Lebesgue Integral? So $ \lim_{t \rightarrow b} \int_a^t f(x) \, dx = \int_{[a,b)} f(x) \, dx . $. I think this holds in the case when $f$ is a nonnegative function as we can apply MCT or DCT, but I can't really figure it out for general case. $\endgroup$ – CL. Jun 28 '17 at 12:33
  • $\begingroup$ Yes, one can use DCT: $f \cdot 1_{[a,t)}$ is dominated by $\lvert f \rvert$ and converges to $f$ pointwise (if necessary, prove it for a sequenc $t_n \uparrow b$ and use that a function converges to a value at a point if and only if it converges on every sequence with limit that point). $\endgroup$ – Chappers Jun 28 '17 at 12:44
  • $\begingroup$ The problem with applying DCT with the given definition is we need $\int |f| \, d\mu < \infty$ ($\mu$ is Lebesgue measure.) If $f$ is improperly integrable, then we can, as you said, take an arbitrary sequence $t_n \rightarrow b$, so that $\lim_{ n \rightarrow \infty} \int_a^b f_n \, dx =: \int_a^b f \, dx $ (definition), where $f_n \rightarrow f$. BUT the problem is we cannot apply DCT or MCT. So we cannot say $\int_{[a,b) } f \, d\mu = \int_a^b f (x) \, dx $ right? $\endgroup$ – CL. Jun 28 '17 at 21:37
  • $\begingroup$ If $f$ is improperly Riemann integrable $\int_a^b f(x) \, dx $ is a Lebesgue integral if and only if $\int_a^b \lvert f(x) \rvert \, dx < \infty$ (due to how the Lebesgue integral is defined, $f$ Lebesgue-integrable requires $\lvert f \rvert$ to be as well). $\endgroup$ – Chappers Jun 29 '17 at 5:29

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