Let $(\mathbf{R}, M, m)$ be Lebesgue measure space and $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function. Show that

$$f \in L^1_m (\mathbf{R}) \text{ if and only if } \mathop{\lim_{a \to -\infty}}_{ b \to +\infty} \int_a^b |f(x)| \,\mathrm{d} x \text{ exists and it is finite.}$$

Moreover, $$\int_\mathbf{R} |f(x)| \,\mathrm{d} m = \mathop{\lim_{a \to -\infty}}_{ b \to +\infty} \int_a^b |f(x)| \,\mathrm{d} x \text{ and } \int_\mathbf{R} f(x) \,\mathrm{d} m = \mathop{\lim_{a \to -\infty}}_{ b \to +\infty} \int_a^b f(x) \,\mathrm{d} x.$$

  • $\begingroup$ I don't understand the downvote. $\endgroup$ – Karatug Ozan Bircan Jan 19 '13 at 16:25
  • $\begingroup$ I understand how one could downvote this. It is only a statement of an exercise in measure theory with no information on what the poster did to solve the problem, or any other thoughts, no context of where the problem was encountered, etc. The description on the downvote arrow starts with "This question does not show any research effort." Although the poster simultaneously posted an answer, showing that there wasn't a lack of effort involved, looking just at the quality of the question itself leaves something to be desired. (I am not downvoting and don't presume to know why someone did.) $\endgroup$ – Jonas Meyer Jan 19 '13 at 18:26
  • $\begingroup$ @JonasMeyer Well, it's OK to ask and answer our own questions (See blog.stackoverflow.com/2011/07/…). That is why I posted an answer as well. $\endgroup$ – Karatug Ozan Bircan Jan 19 '13 at 18:29
  • 1
    $\begingroup$ Thank you. I was aware of that. It does not seem relevant to what I stated as the reason I could understand downvoting (i.e., lack of research effort, etc., shown in the question itself), so perhaps my earlier comment was unclear. To me, the fact that you also posted an answer makes it clear that you did work on the problem more than the question post indicates. But questions can be judged on their own terms as well. $\endgroup$ – Jonas Meyer Jan 19 '13 at 18:36

$\Longrightarrow:$ Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function. Assume that $f \in L^1 (\mathbf{R})$. Then $\int_\mathbf{R} |f| dm < \infty$ by definition. Also we know that every continuous function on the interval $[a,b]$ is Riemann integrable. Hence since $|f|$ is continuous on $[a,b]$, $|f|$ is Riemann integrable on the interval $[a,b]$. We also know that $$\int_{[a,b]} |f| dm = \int_a^b |f| dx.$$

Also since $|f|$ is non-negative, we obtain the following the inequality

$$\int_{[a,b]} |f| dm \leq \int_\mathbf{R} |f| dm \text{ for any } a,b \in \mathbf{R}.$$

Note that the value of the integral is increasing. Therefore by Monotone Convergence Theorem, we have:

$$\lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_a^b |f(x)| dx=\lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_{[a,b]} |f(x)| dm = \int_\mathbf{R} |f| dm.$$

$\Longleftarrow:$ On the other hand, suppose that the limit $\lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_a^b |f(x)| dx$ exists and it is finite. We know that $$\int_{[a,b]} |f| dm = \int_a^b |f| dx.$$

Taking the limit of both sides, we obtain

$$\lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_a^b |f(x)| dx=\lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_{[a,b]} |f(x)| dm= \int_\mathbf{R} |f| dm,$$

as desired.

In order to prove, $\int_\mathbf{R} f(x) dm = \lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_a^b f(x) dx$, we write $f$ as a difference of two non-negative functions $f=f^+ - f^-$. Since $f^+$ and $f^-$ are both non-negative, the result we proved holds. Hence $$\int_\mathbf{R} f^+(x) dm = \lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_a^b f^+(x) dx$$ and $$\int_\mathbf{R} f^-(x) dm = \lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_a^b f^-(x) dx$$ and therefore we conclude that $$\int_\mathbf{R} f(x) dm = \lim_{a \rightarrow - \infty, b \rightarrow \infty} \int_a^b f^+(x) - f^-(x) dx,$$ as desired.


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