Let $\mathcal{L}^0$ be the set of all measurable functions, $\mathcal{L}^1$ be the set of all integrable functions in $\mathcal{L}^0$ and $\mathcal{L}^{0-1}$ the set of all functions $f\in \mathcal{L}^0$ with $f^-\in \mathcal{L}^1$
we set $$\int fd\mu=\int f^+d\mu-\int f^-d\mu\in(-\infty,\infty], \text{ for } f\in \mathcal{L}^{0-1}$$
How do I show that the Lebesgue integral remains a monotone operation in $\mathcal{L}^{0-1}$?
I think I need to show:
If $f\in \mathcal{L}^{0-1}$ and $g\in \mathcal{L}^0$ are such that $g(x)\geq f(x)$, for all $x\in S$, then $g\in \mathcal{L}^{0-1}$ and $\int gd\mu\geq \int fd\mu$.
- Is that a correct approach?
- If so how do I show this?
- If not what is a good approach?
Arguably, if one shall prove that the integral is monotone on $\mathcal{L}^{0-1}$, one can assume from the start that both, $f$ and $g$, belong to $\mathcal{L}^{0-1}$ rather than deducing $g \in \mathcal{L}^{0-1}$ from $f \in \mathcal{L}^{0-1},\, g \in \mathcal{L}^0$ and $f \leqslant g$.
There's nothing wrong with doing the latter, though.
Note that for any functions $f,g$ the inequality $f \leqslant g$ implies
\begin{align} f^+ &\leqslant g^+,\text{ and} \tag{1}\\ g^- &\leqslant f^-. \tag{2} \end{align}
For measurable $g$ and $f \in \mathcal{L}^{0-1}$, one can easily deduce $g \in \mathcal{L}^{0-1}$ and
$$\int f\,d\mu \leqslant \int g\,d\mu\tag{3}$$
from $(1)$ and $(2)$.
From $0 \leqslant g^- \leqslant f^-$ and $f^- \in \mathcal{L}^1$ it follows that $g^- \in \mathcal{L}^1$, and hence $g \in \mathcal{L}^{0-1}$.
Adding $(1)$ and $(2)$ and integrating yields
$$\int f^+\,d\mu + \int g^-\,d\mu \leqslant \int g^+\,d\mu + \int f^-\,d\mu\tag{4}$$
by the monotonicity of the Lebesgue integral for non-negative functions. Subtracting the finite value $\int g^-\,d\mu + \int f^-\,d\mu$ from $(4)$ yields $(3)$.