# An unusual definition of the Lebesgue integral

In my analysis class, I was given the following definition of the Lebesgue integral:

Let

$O^b:=\{ \phi:[a,b]\to\mathbb R \mid \exists \text{ sequence of step functions$s_1 \leq s_2 \leq \ ... \ \rightarrow \phi$a.e.}\}$

$O^a:=\{ \phi:[a,b]\to\mathbb R \mid \exists \text{ sequence of step functions$s_1 \geq s_2 \geq \ ... \ \rightarrow \phi$a.e.}\}$

On $O^b$ define $\int^b_a \phi:= \lim \int^b_a s_n \leq +\infty$

On $O^a$ define $\int^b_a \phi:= \lim \int^b_a s_n \geq -\infty$

$L$-$\overline{\int^b_a}f:= \inf \{ \int^b_a \phi \mid f\leq \phi \in O^b \}$

$L$-$\underline{\int^b_a}f:= \sup \{ \int^b_a \phi \mid f\geq \phi \in O^a \}$

$O^b$ and $O^a$ stand for One-sided$^{\text{below}}$ and One-sided$^{\text{above}}.$

If $L$-$\underline{\int^b_a}f = L$-$\overline{\int^b_a}f$, we call it the Lebesgue integral of $f$.

In order to get a better understanding of what's going on, I tried to look up the definition in a textbook or on the internet but I couldn't find anything (not even close to the above). The problem is that we didn't have any measure theory so far: only the zero set (hence the "a.e." above). And all texts seem to define the Lebesgue integral after introducing measure etc.

I would highly appreciate a reference to a material which introduces the Lebesgue integral in the above way.

• Have you tried asking your prof? Surely he/she didn't just make up a new formulation of the theory out of the blue just to confuse everybody. – Jeff Jun 29 '13 at 0:21
• @Jeff: Yes, we did ask him. He gave us a reference to one book yet it turned out to be the same as the rest: first a chapter on measure theory, then the Lebesgue integral. – Leo Jun 29 '13 at 0:28
• Ah, I guess it is not necessary to include it here. Looks clumsy, and it is clear we're talking about Lebesgue. – Pedro Tamaroff Jun 29 '13 at 0:38
• I don't think this gives the Lebesgue integral unless by "step function" you mean simple function. – nullUser Jun 29 '13 at 0:48
• Unless $f$ is bounded this doesn't work. Take $f(x) = x^{-1/2}$ on $[0,1]$. Any step function above $f$ has integral $+\infty$ but of course $\int_0^1f < \infty$. – nullUser Jun 29 '13 at 0:50