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If $f$ is a measurable function on space $X$ with measure $\mu$, we may define $\int_X f d\mu = \sup\{\int_X s d\mu | s\text{ simple measurable on }X, s \le f\}$. But it seems that the right hand side is defined for all $f$ on $X$. So why don't we define $\int_X f d\mu$ as such for all $f$ on $X$?

mr. j
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2 Answers2

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The right hand side is well defined for any nonnegative $f$. But if we want to introduce a useful theory then the operation of integration should have some "nice" properties (such as being linear for example). The system of measurable functions is reasonably large for many applications and does not spoil these "nice" properties.

  • Could we not extend the definition to non-measurable $f$ by defining $\int f = \int f^+ - \int f^-$ using the extended definition of $\int$ (provided we don't get $\infty - \infty$)? That obviously doesn't solve the utility issue. – mr. j Jan 27 '14 at 20:23
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    Yes, we could. But if you define the integral for every function in this way, then one could find e.g. two nonnegative functions $f,g$ such that $\int f=\int g=0$ but $\int (f+g)=1$. This is the reason why we restrict ourselves only to measurable functions - then there are no such patological examples. –  Jan 27 '14 at 20:29
  • It seems to me that the more natural definition of $\int$ is the general one but that theorems that require certain properties like linearity should restrict themselves to measurable functions. Perhaps some theory would hold in the general case. But I suppose it's probably more convenient to restrict at the outset. – mr. j Jan 27 '14 at 20:59
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There is nothing intrinsically wrong with this definition for all $f$. The point of Lebesgue's theory, however, is that measurable functions (and measurable functions alone) are well approximated by simple functions.

Thus, you get nice theorems like Monotone/Dominated Convergence, which won't hold in general for any function.

L..
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  • I don't understand what you mean by measurable functions alone being well approximated by simple functions. If $\int f = \sup{ \int s }$, doesn't that mean that there is a sequence $s_n$ with $\int s_n \rightarrow \int f$, and thus the simple measurable functions well-approximate $f$ (in the L1 sense) whether or not $f$ is measurable? – mr. j Jan 27 '14 at 20:39
  • @mr.j one instance of what I mean is this: if $f$ is measurable, we can find a sequence of simple measurable functions $\phi_n$ converging pointwise and monotonically to $f$. This is very useful when you combine it with the Monotone Convergence Theorem (because then $\int \phi_n$ converges to $\int f$). If $f$ is not measurable, you could still find the functions $\phi_n$, but they won't be integrable. So this is stronger than just approximating the integrals, like you stated. – L.. Jan 27 '14 at 20:51