I am trying to understand why the Lebesgue integral is defined using the lower Lebesgue sum rather than the upper Lebesgue sum. I am trying to prove the following inter-related propositions:
(a) Suppose $(X, \mathcal S, \mu)$ is a measure space with $\mu(X)<\infty$. Suppose that $f\colon X\to[0, \infty)$ is a bounded $\mathcal S$-measurable function. Prove that
\begin{align*}
\int f \; d\mu = \inf\left\{\sum_{j=1}^m \mu(A_j) \sup_{A_j} f: A_1, \ldots, A_m \text{ is an } \mathcal S\text{-partition of } X \right\}
\end{align*}
The author of my text calls the expression on the right hand side of the equation above the upper Lebesgue sum. My textbook describes the Lebesgue integral as: $\int f \; d\mu = \sup\left\{\sum_{j=1}^m \mu(A_j) \inf_{A_j} f: A_1, \ldots, A_m \text{ is an } \mathcal S\text{-partition of } X \right\}$, so I know that I have to show that $$\sup\left\{\sum_{j=1}^m \mu(A_j) \inf_{A_j} f: A_1, \ldots, A_m \text{ is an } \mathcal S\text{-partition of } X \right\} = \inf\left\{\sum_{j=1}^m \mu(A_j) \sup_{A_j} f: A_1, \ldots, A_m \text{ is an } \mathcal S\text{-partition of } X \right\}$$ I'd really appreciate it if someone can show me how this can be done.
(b) Show that the conclusion of part (a) can fail if the condition that $\mu(X)< \infty$ is deleted.
I have no clue how to approach this one. Can someone give me some examples of sets whose Lebesgue measure is infinite? I'll try to complete the solution from there.
Any help on one or both of these questions would be greatly appreciated!
Best Answer
Many details omitted:
Suppose $B$ is an upper bound for $f$. Then $B-f \ge 0$. Note that $B = B-f +f$ and $\int B = \int (B-f) + \int f$.
\begin{eqnarray} \int (B-f) &=& \sup \{ \sum_k (\inf_{A_k} (B-f)) \mu A_k \} \\ &=& \sup \{ \sum_k (B-\sup_{A_k} f) \mu A_k \} \\ &=& \sum_k B \mu A_k +\sup \{ \sum_k (-\sup_{A_k} f) \mu A_k \} \\ &=& \int B -\inf \{ \sum_k (\sup_{A_k} f) \mu A_k \} \\ \end{eqnarray} Now compare the two equations.
For part (b) consider $f(x) = {1 \over x^2}$ on $x \ge 1$.