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)=1knSup{f(x):x[tk1,tk]}(tktk1)L(f,P)=1kninf{f(x):x[tk1,tk]}(tktk1)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)f(x)? If sup{f(x):x[a,b]}=λ\operatorname{sup}\left\{f(x): x\in[a,b]\right\}=\lambda but then f(x)f(x) is constant so infimum of f(x)f(x) is also λ\lambda which immediately follows that L(f,P)=λ(ba)=U(f,P)L(f,P)=\lambda(b-a)=U(f,P) Further L(f)L(f,P),  U(f)U(f,P)    L(f)=U(f)=λ(ba)L(f)\geq L(f,P) ,\; U(f)\leq U(f,P) \implies L(f)=U(f)=\lambda(b-a) shows that f(x)f(x) is integrable and its values isL(f)abf(x)U(f)    abf(x)dx=λ(ba)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.

Note by Naren Bhandari
1 year ago

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