If $f$ is non-negative and improperly integrable on $\mathbb{R}$, prove that $f$ is Lebesgue integrable.

The way I thought about this is that if $f(x)$ is improperly integrable and non-negative then $ \int f \, dx = \int \lvert f \rvert \, dx$ which implies Lebesgue integrability.

I feel like this isn't correct though. Also what is a definition of improper integrability on $[0, \infty)$? The only definition in my book looks at from $(a+ \epsilon, b]$ as $\epsilon \to 0$.


Assume that $f$ is non-negative and Riemann integrable over $[0,c]$ for all $c > 0$. The Riemann and Lebesgue integrals are equivalent on the bounded interval.

Hence, $$\int_{[0, \infty)} f = \lim_{c \to \infty} \int_{[0,\infty)} f \chi_{[0,c]} = \lim_{c \to \infty}\int_{[0,c]} f = \lim_{c \to \infty}\int_0^cf(x) \, dx = \int_0^\infty f(x) \, dx$$

where the left limit follows from the MCT for Lebesgue integrals and the right limit is a consequence of improper integrability of $f$.

The proof is similar for a Type I improper integral on a bounded interval.

  • $\begingroup$ So the MCT is what implies $$\int_{[0, \infty)} f = \lim_{c \to \infty} \int_{[0,\infty)} f$$? $\endgroup$
    – MathNoob
    Apr 8 '17 at 2:41
  • $\begingroup$ You left out the all important function $\chi_{[0,c]}$ which equals $1$ if $0 \leqslant x \leqslant c$ and $0$ otherwise. Hence $f(x) \chi_{[0,c]}(x)$ is increasing (since $f$ is non-negative) and convergent to $f(x)$ as $c \to \infty$ for every $x$. Then we can apply MCT and we know that the integral of $f \chi$ over $[0,\infty)$ exists since it is the same as the integral over $[0,c]$ by Riemann integrability. $\endgroup$
    – RRL
    Apr 8 '17 at 2:48

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