Let (X,$\Lambda$,$\mu$) be a measure space such that $\mu(X)$ is finite and let $M^r=M^r(X,\mu)$ be the set of all real $\mu$ almost everywhere finite valued $\mu$-measurable functions on $X$ with identification of $\mu$-almost equal functions.
Assume $(f_t:t\in T)$ is an upwards directed set of positive elements of $M^r(X,\mu)$ with the upper bound $g$, i.e., $0\leq f_t\uparrow\leq g$.
For $P=\sup\{\int_Xf_td\mu:t\in T\}$, we can choose an increasing sequence $(f_{_{t_n}}:n\in \Bbb{N})$ in the set $(f_t:t\in T)$ such that $$\int_X f_{_{t_n}}d\mu\uparrow P, \text{as}\hspace{0.5cm} n\rightarrow \infty.$$
The pointwise supremum $F(x)=\lim_{n\rightarrow \infty}f_{_{t_n}}(x)$ is then a $\mu$-measurable function and $$\int_XFd\mu=P.$$
Now here is my question,
Question: How can we show that for every $t\in T$, $$f_t(x)\leq F(x),$$ for almost every $x$ in $X$ and therefore $F(x)=\sup(f_t(x):t\in T)$ in $M^r(X,\mu)$?
Many thanks