I am studying real analysis from the book which is written by Manfred Stoll and somewhere in Lebesgue integral properities(page:476) I stuck. I don't believe totally prove the property so I will be happy if you could tell me where i did wrong. Thanks in advance
Proposition:
$A \subset \mathbb{R}$ is a measurable bounded set and $f,g :A \rightarrow \mathbb{R}$ are measurable and non-negative functions. If $f \leq g$ $ \,$a.e. on the set A then
$\int_A f d \lambda \leq \int_A g d \lambda$ $\quad$ holds.
My attempt:
I have the definition about Lebesgue integrability and (proved) lemma which i used for proof are in below;
Definition : Let $A \subset \mathbb{R}$ bounded and measurable set.
$f:A \rightarrow \mathbb{R}$ non-negative real valued function.
Define , $f_n(x)= min\{f(x),n\}$ then,
$\int_A f d \lambda = lim_{n \rightarrow \infty} \int_A f_n d\lambda$
Lemma*: $ f: [a,b] \rightarrow \mathbb{R}$ is bounded and measurable function and $f \leq g$ a.e. on [a,b] then
$\int_{[a,b]} f d \lambda \leq \int_{[a,b]} g d \lambda$
Proof(of the proposition): Let $ f \leq g$ $\,$ a.e on the set bounded A. Then there exits $a,b \in \mathbb{R}$ such that $A \subset [a,b]$. We can define $g_{x_{A}}$ and $f_{x_{A}}$ as ;
$f_{x_{A}}= \begin{cases} f(x) & x \in A\\ 0 & x \notin A \end{cases}$
and
$g_{x_{A}}= \begin{cases} g(x) & x \in A\\ 0 & x \notin A \end{cases}$
Now we will define the function sequences
$f_n ,g_n :[a,b] \rightarrow \mathbb{R}$ as;
$f_n(x)= min\{f(x),n\}$ $\quad $ and $\quad$ $g_n(x)= min\{g(x),n\}$
So by using Lemma*, we can say that $f_n$ and $g_n$ function sequences are measurable $ \forall n \in \mathbb{N}$ and $ f_n \leq g_n$ $\,$ a.e on [a,b] so that
$\int_{[a,b]} f_n d \lambda \leq \int_{[a,b]} g_n d \lambda$ $\iff$ $\int_A f_n d \lambda \leq \int_A g_n d \lambda$
Now take limit from both sides of inequility,
$\int_A f d \lambda = lim_{n \to \infty}\int_A f_n d \lambda \leq lim_{n \to \infty}\int_A g_n d \lambda =\int_A g d \lambda$