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Non-mathematician here, trying to grasp theory of integration.

Why is it that the integral (or the $\mu$-integral) in, say, a measure theory book, is defined for measurable functions? The definition itself is first given for a simple function, which makes sense, and then we define the general integral of a positive integrand $g$ as $$ \sup\{ \int f \ d \mu : f \le g, f \ \text{is simple and positive}\}$$ and then further we define the integral of any $g$ in terms of $g^+$ and $g^-$ which are positive.

But $g$ is always required to be measurable...where does measurability come into play here? Why is that important? Where does the theory fall apart when $g$ is non-measurable?

Chappers
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  • @StefanHansen: That's a restriction on $f$, not on $g$. That $f$ needs do be simple doesn't translate to $g$ needing to be simple (otherwise the definition would be pointless). Note that non-measurable sets $S$ have measurable subsets (like the empty set, and single-element sets for each element of $S$). Given the completeness of real numbers, the only possibility I see for the supremum not to exist is if the set is unbounded. – celtschk Sep 21 '15 at 10:13
  • Intuitively, what the definition is attempting to do is inscribing simple functions beneath the graph of a positive function $g$, and using the area beneath the simple functions to approximate the area (i.e, integral) of $g$, via the supremum operator. So I guess my question is: Why will this not work if $g$ is non-measurable? Can this approximative process not be done for non-measurable functions? – Chappers Sep 21 '15 at 10:31

2 Answers2

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When you calculate a Riemann integral, you usually take an infinitesimal step along the "$x$"-axis and approximate the function $f$ on this interval as a constant or as a line (rectangular or trapezoidal approximation).

In Lebesgue integration, you do a different thing. You take infinitesimal steps along the "$y$"-axis and ask: For which $x$ is the function $f$ equal to $y$? This gives you a set $S(y)$, the preimage of $y$ under $f$; to be more precise,

$$ S(y) = f^{-1}(y)= \{x{:}\ y\le f(x)\le y+dy\}.$$

Integration now means that you multiply the "size" of this set (its measure) by $f(x)$, and sum the results for all $y$. But the set $S(y)$ can only be measured if the function $f$ is measurable. That's where this term comes in. In other words, if there is an $y$ for which the set $S(y)$ is not measurable, there is a term missing in the computation of the integral.

Bernhard
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  • This is correct, and it furnishes an alternate definition of the Lebesgue integral (when some details are ironed out, of course). I'm not sure it explains why the definition in terms of simple functions doesn't give a valid definition of the integral of the indicator function of the Vitali set (or some other nonnegative nonmeasurable function). – Ian Sep 21 '15 at 12:06
  • Hmm, this is correct, simple functions are by definition measureable. That's why I prefer this other definition of the Lebesgue integral, because it shows why a measurable function is needed in an intuitive way... – Bernhard Sep 21 '15 at 13:10
  • The "partition" perspective is much more intuitive, I agree. But the "simple function approximation" approach has advantages: for instance, independence of the integral on the sequence used for its approximation is essentially "free". One way of unifying the two perspectives is to note that the "partition" definition of the Lebesgue integral is a special case of the "simple function" definition. – Ian Sep 21 '15 at 15:26
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    Specifically, if you define $I(f)=\lim_{n \to \infty} n m(f^{-1}([n,\infty))) + \sum_{i=1}^{n^2} f(i/n) m(f^{-1}([(i-1)/n,i/n]))$, then $I(f)$ is the limit of the integrals of the simple functions $s_n=n \chi_{f^{-1}([n,\infty))} + \sum_{i=1}^{n^2} f(i/n) \chi_{(f^{-1}([(i-1)/n,i/n])}$. – Ian Sep 21 '15 at 15:28
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I have two perspectives to suggest.

Given a function $g$, let

$$I_+(g) = \inf \left \{ \int f d \mu : f \text{ is simple },f \geq g \right \} \\ I_-(g) = \sup \left \{ \int f d \mu : f \text{ is simple },f \leq g \right \}.$$

Then $g$ is bounded and measurable if and only if $I_+(g)=I_-(g)$. This parallels the Riemann situation (replacing piecewise constant functions by simple functions). You can extend to the unbounded case after you have already developed the bounded case.

Alternately, for nonnegative $g$, define $\int g d \mu$ to be $\lim_{n \to \infty} \int s_n d \mu$, where $s_n$ is an increasing sequence of simple functions which converges a.e. to $g$. Then such a sequence exists if and only if $g$ is measurable, and the value obtained is independent of the sequence of simple functions chosen.

The first perspective is closer to how you've formulated your question, but I think the theory is easier to develop from the second perspective.

Ian
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