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I am studying the Lebesgue integration theory and I am encountered with the definition of the Lebesgue integrability.

First, I will assume $f:X\to\mathbb{R}$ is a $\mathcal{A}$-measurable function for a measure space $(X,\mathcal{A}, \mu)$.

My textbook (written by Richard F. Bass) defines the Lebesgue integration first for all positive simple (hence $\mathcal{A}$-measurable) functions $s$, second for all positive $\mathcal{A}$-measurable functions $f$ as $\int f=\sup_{0\leq s\leq f}\int s$, and generally for all other $\mathcal{A}$-measurable functions $f$ as $\int f=\int f^+-\int f^-$.

So, it means when calculating Lebesgue integration for a given function $f$, we treat the domain of $f$ as two parts, which has the function value of positive and negative, respectively.

Next, the author define the integrability for $f$ as $\int\vert f\vert d\mu<\infty$.

Here, I have a question. I think this definition is come because of $\int f<\infty$ iff $\int f^+,\int f^-<\infty$ by the definition of $\int f$ which we can see from above, and moreover, $\int \vert f \vert<\infty$ implies $\int f^+,\int f^-<\infty$.

Am I correct? Or is there other hidden meaning that I missed?

In addition, I can not understand fully why the Lebesgue integration is defined step-by-step as well as from positive simple functions, to positive functions, finally other functions. In particular, why don't we define the integration for 'negative' or composed(somewhere positive but otherwhere negative) simple functions that I think it could be well-defined?

Please comment me about this ambiguous definition for me. Thanks.

Analysis
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    Consider $f(x)=\sin x.$ Then it is $\int f^+=\infty$ and $\int f^-=\infty.$ If you allow $\int |f|=\infty,$ what would the definition of $\int \sin x=\infty-\infty?$ – mfl Oct 13 '14 at 20:07
  • Some authors distinguish between "integrable" and "of finite integral"; the former means that either $\int f^+ < \infty$ or $\int f^- < \infty$ and the latter means that both are finite. In both cases, $\int f = \int f^+ - \int f^-$ is well defined. – William M. Apr 14 '17 at 01:40

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