Sup and inf of constant function

Problem: Show that the constant function is integrable and find its value of integration.

Suppose \(f:\mathbb [a,b]\to \mathbb R\) such that \(f(x)=\lambda\) where \(\lambda\) is any constant. Let \(P\) be any partition on \([a,b]\), ie \[P=\left\{a=t_0<t_1<t_2\cdots< t_n=b\right\}\] then Upper Darboux sum and Lower Darboux sum we evaluate by $U(f,P)=\sum_{1\leq k\leq n}\operatorname{Sup}\left\{f(x): x\in [t_{k-1},t_k]\right\}(t_k-t_{k-1})\\ L(f,P)=\sum_{1\leq k\leq n}\operatorname{inf}\left\{f(x): x\in [t_{k-1},t_k]\right\}(t_k-t_{k-1})$ Now what about the supremum and infimum of $f(x)$? If $\operatorname{sup}\left\{f(x): x\in[a,b]\right\}=\lambda$ but then $f(x)$ is constant so infimum of $f(x)$ is also $\lambda$ which immediately follows that $L(f,P)=\lambda(b-a)=U(f,P)$ Further $L(f)\geq L(f,P) ,\; U(f)\leq U(f,P) \implies L(f)=U(f)=\lambda(b-a)$ shows that $f(x)$ is integrable and its values is$L(f)\leq \int_a^b f(x) \leq U(f)\implies \int_a^b f(x) dx =\lambda(b-a)$

Now how to show that the supremum and infimum of the constant function is constant itself without using completeness property?

Any sorts of help will be appreciated.