# Why Not To Define the Lebesgue Integral Using the Upper Lebesgue Sum.

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!

• Actually, the approach of Daniell-Stone uses something similar to upper and lower intergrals. Commented May 6, 2021 at 3:56
• @OliverDiaz Unfortunately, I am not familiar with that representation. It might be beyond the scope of the text I am reading. Commented May 6, 2021 at 4:01
• I can imagine... Any way, what you have are sums of the form $g=\sum^n_{k=1}a_k\mathbb{1}_{A_j}$ and $G=\sum^n_{k=1}b_k\mathbb{1}_{A_j}$ where $a_j\leq f\leq b_j$ in $A_j$. The rest is monotone convergence. A typical partition is $\{2^{-n}k<f<2^{-n}(k+1)$ for $0\leq k\leq n2^n$ if $f$ is bounded. and nonnegative Commented May 6, 2021 at 4:05
• Its not clear what $(a)$ is, but I suspect this link gives a counterexample math.stackexchange.com/a/2957125/80734 Commented May 6, 2021 at 5:00
• @CalvinKhor Could you please explain what part of (a) is not clear? Commented May 6, 2021 at 5:05

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 $$x \ge 1$$.
• Can you provide some hints on how we can show that $\inf\left\{\sum_{j=1}^\infty \mu(A_j) \sup_{A_j} f: A_1, \ldots\text{ is an } \mathcal S\text{-partition of } X \right\} \ne 1$ where $f$ is the function you suggested for (b)? What even is the value of $\inf\left\{\sum_{j=1}^\infty \mu(A_j) \sup_{A_j} f: A_1, \ldots\text{ is an } \mathcal S\text{-partition of } X \right\}$? Commented May 7, 2021 at 21:34
• Since the partition is finite, at least one $A_k$ has infinite measure, and $\sup_{A_k} f$ is strictly positive, hence the $\inf (\cdots) = \infty$ even though you can evalute the integral to get one. Commented May 7, 2021 at 21:49
• I didn't know as such, however it is a trick used in some proofs when you need non negativity and you have an upper bound, here the quantity is $B-f$ which is non negative. Then I just applied your integral definition to $B-f$ and ground my way through the details. Good luck! Commented May 7, 2021 at 22:33