Let $f:[a,b]\to\mathbb R$ be a bounded nonnegative function. Prove that
$$
\underline{\int_a^b} f(x)dx=\sup_{\xi}\int_a^b \xi(x)dx,
$$
where $\xi$ belongs to the set of step-functions such that $\xi(x)\leq f(x)$ for all $x\in[a,b]$.Show that an analogous result holds if we take $\xi$ continuous or $\xi$ integrable (keeping the hypothesis $\xi(x)\leq f(x)$ for all $x\in[a,b]$).
Let $X$ be the set of step-functions $\xi:[a,b]\to\mathbb R$ such that $\xi(x)\leq f(x)$ for all $x\in[a,b]$. Take $P=\{a=t_0,t_1,\ldots,t_{n-1},t_n=b\}$ be a partition of $[a,b]$. So
$$
\underline{\int_a^b} f(x)dx = \sup_{P} s(f;P),
$$
where $s(f;P)=\sum_{i=1}^n m_i(f)(t_i-t_{i-1})$ and $m_i(f)=\underset{x\in[t_{i-1},t_i]}{inf} f(x)$.
Let $\xi\in X$. From $\xi(x)\leq f(x)$ for all $x\in[a,b]$, we get
$$
\int_a^b \xi(x)dx=\underline{\int_a^b} \xi(x)dx\leq \underline{\int_a^b} f(x)dx \Longrightarrow \underset{\xi}{sup} \int_a^b \xi(x)dx \leq \underline{\int_a^b} f(x)dx.
$$
Define $\xi_1:[a,b]\to\mathbb R$ by $\xi_1(x)=m_i(f)$ for all $x\in[t_{i-1},t_i]$. Hence,
$$
\int_a^b \xi_1(x)dx = \sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) \Longrightarrow \int_a^b \xi_1(x)dx = \sup_{P} s(f;P)
$$
Thus the supremum is attained. Therefore,
$$
\underset{\xi}{sup} \int_a^b \xi(x)dx = \underline{\int_a^b} f(x)dx.
$$
Is my proof correct? If not, why?
Best Answer
You are close to being correct.
The function $\xi_1$ you introduce belongs to $X_1$, the collection of step functions dominated by $f$ that correspond to lower Riemann sums. You are correct in writing
$$\int_a^b \xi_1(x)dx = \sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) $$
but not correct in claiming attainment of the supremum by a step fuction through the subsequent implication
$$\int_a^b \xi_1(x)dx = \sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) \Longrightarrow \int_a^b \xi_1(x)dx = \sup_{P} s(f;P) $$
The equality $\int_a^b \xi_1(x)dx = s(f;P)$ for a specific partition does not imply $\int_a^b \xi_1(x)dx = \sup_{P} s(f;P)$.
What is true, however, is that, since there is a one-to-one correspondence between functions $\xi_1 \in X_1$ and lower sums,
$$\sup_{\xi_1 \in X_1}\int_a^b \xi_1(x)dx = \sup_{P} s(f;P) = \underline{\int_a^b} f(x)dx$$
Since $X_1 \subset X$ it follows that
$$\left\{\int_a^b \xi_1(x) \, dx \,\,|\,\, \xi_1 \in X_1\right\} \subset \left\{\int_a^b \xi(x) \, dx \,\,|\,\, \xi \in X\right\},$$
and
$$\tag{*}\underline{\int_a^b} f(x)\,dx = \sup_{\xi_1 \in X_1} \int_a^b \xi_1(x) \, dx \leqslant \sup_{\xi \in X} \int_a^b \xi(x) \, dx \leqslant \underline{\int_a^b} f(x)\,dx $$
Recall that you correctly proved the far right inequality in (*), and we have as desired,
$$\underline{\int_a^b} f(x)\,dx = \sup_{\xi \in X} \int_a^b \xi(x) \, dx $$