An alternative definition of the Lebesgue integral

Question

I am asking this in the context of improper Riemann integrals existing but Lebesgue integrals don't. The sinc function has improper Riemann integral but fails to be Lebesgue integrable. Capinski and Kopp repeatedly keep emphasizing that Lebesgue integral has an absolute nature because Lebesgue integral exists if and only if the positive and negative parts of the function have Lebesgue integral (which by definition must be finite). This is also in line with the fact that Lebesgue integral exists if and only if the absolute value of the function has a Lebesgue integral.

The Lebesgue integral for non-negative functions is defined

Lebesgue Integral for non-negative Real Functions: For any non-negative measurable function $f$, and a measurable set $E \in \mathcal{M}$, the Lebesgue integral is defined as

$$\int_E fdm = \sup Y(E,f)$$

where $$Y(E,f) = \left \{ \int_E \varphi dm : 0 \leq \varphi \leq f,~\varphi ~ \text{simple} \right\}.$$

Here, $(\mathbb{R},\mathcal{M},m)$ represents the Lebesgue measure space.

But when extending the definition to a general function (so not a non-negative or non-positive function alone but any function), they define it as

Definition of Lebesgue Integral of any Real Function: If $E \in \mathcal{M}$ and the measurable function $f$ has both $\int_E f^+ dm$ and $\int_E f^- dm$ finite, then we say that $f$ is integrable, and define $$\int_E fdm = \int_E f^+dm − \int_E f^-dm.$$

This is where the problem lies. This definition gives rise to the constraint that the function be absolutely Lebesgue integrable.

Suppose, we defined the Lebesgue integral of a function which is not necessarily non-negative or non-positive, as being the same as for the case when it was defined for the non-negative case, then we would be allowing cancellations thereby letting more functions Lebesgue integrable. Specifically, we could define Lebesgue integral of a bounded function as the supremum of the integrals of simple functions bounded above by the given function.

Why then did the founding fathers define it differently? Specifically, what is the problem in the following definition of a Lebesgue integral of a bounded function (but not necessarily non-negative or non-positive)?

Alternative Definition of the Lebesgue Integral: For any bounded and measurable function $f$, and a measurable set $E$, the Lebesgue integral is defined as

$$\int_E fdm = \sup Y(E,f)$$

where $$Y(E,f) = \left \{ \int_E \varphi dm : \varphi \leq f,~\varphi ~ \text{simple} \right\}.$$

Lebesgue Integral for Simple Functions:

Let the simple function $\varphi(x) \triangleq \sum_{i=1}^n a_i I_{A_i}(x)$, where $A_i \in \mathcal{M}$.

The Lebesgue integral over $E \in \mathcal{M}$ of the simple function $\varphi$ is given by

$$\int_E \varphi dm = \sum_{i=1}^n a_i m(A_i \cap E).$$

Obviously, I am now including a big constraint that the function be bounded. But this does have the benefit of including many functions that are otherwise not integrable when defined in the traditional way. So why exactly was it defined in such a way that the Lebesgue integral has this absolute integrability constraint?

Answer 1

With your definition, the following holds:

Let $f:\Bbb R\to\Bbb R$ be a (bounded) measurable function such that $m(f^{-1}(-\infty,0))=\infty$. Then $\int_{\Bbb R} f\,dm=-\infty$.

Proof: Let $\varphi\le f$ be a simple function and let $\varphi(\Bbb R)=\{a_1,\cdots, a_k\}$. Since $f^{-1}(-\infty,0)$ has infinite measure and $\varphi\le f$, there is some $a_j$ such that $m(\varphi^{-1}(a_j))=\infty$ and $a_j<0$. If $\sum_{u=1}^k a_u m(\varphi^{-1}(a_u))$ is not defined in $[-\infty,\infty]$, then $\varphi$ does not count among the simple functions that determine the elements in $Y(\Bbb R,f)$. However, since one of the terms of $\sum_{u=1}^k a_u m(\varphi^{-1}(a_u))$ is $a_j m(\varphi^{-1}(a_j))=-\infty$, the sum can only exist if all its positive terms are finite, and therefore $\sum_{u=1}^k a_u m(\varphi^{-1}(a_u))=-\infty$. So $Y(\Bbb R,f)\subseteq\{-\infty\}$. Therefore $\int_{\Bbb R} f\,dm\le-\infty$.

Notice that this applies to a very large class of measurable functions, both $L^1$ in the usual sense (like $-(x^2+1)^{-1}$) and not (like $\sin x$). Incidentally, this applies to all the bounded measurable functions for which the usual Lebesgue integral cannot be defined in $[-\infty,\infty]$, like for instance $\frac1x\sin x$.