Equivalent definition of Lebesgue integral

Question

We define $\int f d \mu=\sup \{\mathcal{L}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$ where $\mathcal{L}(f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \inf _{A_{i}} f$ for an $\mathcal{S}$-partition $P=A_{1}, \ldots, A_{n}$.

I am looking to show that when $\mu(X)< \infty$ and $f$ is bounded then this definition is equivalent to using the upper sum $\mathcal{U}(f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \sup _{A_{i}} f$ with $\int f d \mu=\inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$.

It was easy to prove that $\int f d \mu \leq \inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$ by using the fact that $\mathcal{L}(f, P) \leq \mathcal{U}(f, P)$.

I am now left to prove $\int f d \mu \geq \inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$. I think this can be done by first proving it for when $f$ is simple and then generalizing but I am struggling. Any help would be much appreciated. Thanks in advance

Answer 1

Note that $\mathcal{U}(f,P) = \sum_{i=1}^n \sup_{A_i}f\cdot \mu(A_i) = \int\psi_P$ where $\psi_P = \sum_{i=1}^n \sup_{A_i}f\cdot \chi_{A_i}$ is a simple function.

Since $\psi_P \geqslant f$, we have $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \leqslant \mathcal{U}(f,P)$ and, it follows that taking the infimum over all partitions $P$, we have $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \leqslant \inf_P\mathcal{U}(f,P)$.

On the other hand, if $\psi = \sum_{j=1}^m b_j \chi_{B_j}$ is the canonical representation of a simple function $\psi \geqslant f$, then $b_j \geqslant \sup_{B_j} f$ and $\int \psi = \sum_{j=1}^m b_j \mu(B_j) \geqslant \sum_{j=1}^m \sup_{B_j} f\cdot \mu(B_j) = \mathcal {U}(f,P_\psi) $ for some partition $P_\psi$. This implies that $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \geqslant \inf_P \mathcal{U}(f,P).$

Hence, $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} = \inf_P \mathcal{U}(f,P)$ and by a similar argument we also have $\sup \{\int\varphi:\varphi \leqslant f, \varphi \text{ simple} \} = \sup_P \mathcal{L}(f,P)$.

Since $f$ is bounded, for any $\epsilon>0$ there exists, by the simple approximation lemma, simple functions $\varphi_\epsilon$ and $\psi_\epsilon$ such that $\varphi_\epsilon \leqslant f \leqslant \psi_\epsilon$ and $\psi_\epsilon - \varphi_\epsilon < \epsilon$.

Thus,

$$\int \varphi_\epsilon \leqslant \sup \{\int\varphi:\varphi \leqslant f, \varphi \text{ simple} \} = \sup_P \mathcal{L}(f,P) \\ \leqslant \inf_P \mathcal{U}(f,P) = \inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \leqslant \int\psi_\epsilon,$$

and, for all $\epsilon > 0$,

$$0 \leqslant \inf_P \mathcal{U}(f,P) -\sup_P \mathcal{L}(f,P) \leqslant \int \psi_\epsilon - \int \varphi_\epsilon < \epsilon \mu(X)$$

Therefore,

$$\inf_P \mathcal{U}(f,P) = \sup_P \mathcal{L}(f,P) = \int f $$

Answer 2

First of all, the definition of the Lebesgue integral above won't work in general for negative (measurable) functions. Consider $f\colon \mathbb{R} \to \mathbb{R}; f(x) = -e^{-x^2}$. We have $$ - \infty < \int_\mathbb{R} f d\lambda\,,$$ where $\int f d \lambda$ denotes the usual Lebesgue integral wrt. to the Lebesgue measure. (It's usualy defined by approximating measurable $f$ by simple functions. I'd suggest looking this up, since it is rather poorly explained in the book you mentioned.) On the other hand $$ \sup \{\mathcal{L}(f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\} = - \infty $$ since $\mathcal{L}(f, P) = -\infty$ for all Partitions $P$.

Now, in order to use the definition 3.3 from Axler for $\textit{measurable non-negative}$ functions and show the result for measurable bounded $f \colon X \to [0, \infty]$ with $\mu(X) < \infty$, we first use linearity of the integral $$ \int_X f d \mu = -\int_X (-f) d \mu = - \int_X \|f\|_\infty - \|f\|_\infty -f d\mu = - \int_X \|f\|_\infty -f d\mu + \|f\|_\infty \mu(X)\,, $$ where $\|f\|_\infty$ denotes the sup norm. Since $\|f\|_\infty -f \geq 0$ we get by definition 3.3 $$ \int_X f d \mu = -\sup \{\mathcal{L}(\|f\|_\infty -f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\,. $$ Observe that $\mathcal{L}(-f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \inf_{A_{i}} -f = -\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \sup_{A_{i}} f = -\, \mathcal{U}(f,P)$, hence $$ \begin{align} \int_X f d \mu &= - \sup \{-\, \mathcal{U}(f - \|f\|_\infty, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\\ &=\inf \{\, \mathcal{U}(f- \|f\|_\infty, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\,.\\ \end{align} $$ Since $\sup_{A_{i}} (f- \|f\|_\infty) = \sup_{A_{i}} (f)- \|f\|_\infty$ in the defintion of $\mathcal{U}$ it follows that $$ \begin{align} \int_X f d \mu &=\inf \{\, \mathcal{U}(f, P) - \|f\|_\infty \mu(X) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\\ &=\inf \{\, \mathcal{U}(f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}\,.\ \end{align} $$