Properties of lebesgue integral

Question

Supposing we defined the lebesgue integral as

$\int f d \lambda $= sup{$\int s d\lambda$ | $s \leq f $ }

How does this definition lead to the following inequality

$\int (f+g) d\lambda \geq \int f d\lambda + \int g d\lambda$

Answer 1

For a function $f \ge 0$, define the (lower) Lebesgue integral by $$ \int f\;d\lambda = \sup \int s\;d\lambda \tag{1}$$ where the $\sup$ is over all simple functions $s$ with $0 \le s \le f$ everywhere. So it seems we have previously defined $\int s\,d\lambda$. Also, we should have proved some things about it:

$\bullet$ the sum of two simple function is a simple function

$\bullet$ If $s,t$ are simple functions, then $\int(s+t)\;d\lambda = \int s\;d\lambda+ \int t \;d\lambda$

I will write $L(f)$ for the right side of (1) so that I will not inadvertently use unproved properties of the integral.

Given two nonnegative functions $f,g$, we want to prove $L(f+g) \ge L(f)+L(g)$. Let $\alpha \in \mathbb R$ be any number such that $\alpha < L(f)$. Then by the definition of $L(f)$ there is a simple function $s$ with $0 \le s \le f$ and $$ \int s\;d\lambda \ge \alpha $$ Let $\beta \in \mathbb R$ be any number such that $\beta < L(g)$. Then there is a simple function $t$ with $0 \le t \le g$ and $$ \int t\;d\lambda \ge \beta. $$ Note that $s+t$ is a simple function, and $ 0 \le s+t \le f+g$. Therefore by the definition of $L(f+g)$ we have $$ L(f+g) \ge \int (s+t)\;d\lambda $$ Therefore $$ L(f+g) \ge \int s\;d\lambda+ \int t\;d\lambda \ge \alpha+\beta $$ This inequality is true for all $\alpha < L(f)$, so $$ L(f+g) \ge L(f)+\beta $$ This inequality is true for all $\beta < L(g)$, so $$ L(f+g) \ge L(f) + L(g) . \qquad\qquad\square $$

extra note
If we wanted to prove equality, $L(f+g) = L(f)+L(g)$, then we would also need to know: any simple function $u$ with $0 \le u \le f+g$ can be decomposed as $u=s+t$ where $0 \le s \le f$ and $0 \le t \le g$. In general, though, that is false. For example, nonmeasurable functions $f$ and $g$ may fail to work for that.