How to prove equality between two supremums?

Question

$if\ \int_{c_1}^{c_2}f(x)dx\ge0 \ for\ all\ c_1,c_2\ that:\ a\le c_1\le c_1\lt b.\\ So,\ if\ we\ define\ F(t)=\int_{a}^{t}f(x)dx,\ so,\ we\ will\ get\ for\ all\ c\in[a,b]\\F(d)=\int_{a}^{c}f(x)dx+\int_{c}^{d}f(x)dx\ge\int_{c}^{d}f(x)dx\\So,\ we\ get\ that\ sup\{\int_{c}^{d}f(x)dx:[c,d]\subset[a,b)\}=sup\{F(t):t\in[a,b)\}$

How to prove that both sup are indeed the same? The right supremum is indeed the bigger one, but how can I show that he is the smallest compared to all other upper limits of the left expression?

Answer 1

let $\beta=sup\{\int_{c}^{d}f(x)dx:[c,d]\subset[a,b)\}$ and $\alpha=sup\{F(t):t\in[a,b)\}$. Assume $\beta<\alpha$, then $\delta=\alpha-\beta>0$. By the definition of supremum, we can find $t\in[a,b)$ s.t. $F(t)>\alpha-\frac{\delta}{2}$. Since $[a,t]\subset[a,b), F(t)\in \{\int_{c}^{d}f(x)dx:[c,d]\subset[a,b)\}$, so $\beta\ge F(t)>\alpha-\frac{\delta}{2}$. Thus $\frac{\delta}{2}>\alpha-\beta=\delta$ and we have a contradiction. Therefore $\beta\ge\alpha$.

• Remark: I implicitly assumed that $\alpha<\infty$. If $\alpha=\infty$, then there is a sequience $\{F(t_n)\}$ s.t. $F(t_n)\rightarrow\infty$. Since each $F(t_n)\in\{\int_{c}^{d}f(x)dx:[c,d]\subset[a,b)\}$, $\beta=\infty$ as well.