How to prove $L(f, P_{1}) \leq U(f, P_{2})$ under given conditions.

I would like to show that under the assumptions of the following Theorem, if $$P_{1}$$ and $$P_{2}$$ are partitions of $$[a, b]$$ then $$L(f, P_{1}) \leq U(f, P_{2})$$, and I would like to use this result to directly prove the theorem.

Theorem: Suppose that the functions $$f : [a, b] \rightarrow \mathbb{R}$$ and $$g : [a, b] \rightarrow \mathbb{R}$$ are integrable and that

$$f(x) \leq g(x)$$

for all $$x$$ in $$[a, b]$$. Then,

$$\int_{a}^{b} f \leq \int_{a}^{b} g.$$

So, I recognize the above Theorem as the monotonicity of the integral. Also, since $$f$$ is integrable, I tried doing some things with the Archimedes-Riemann theorem, but I didn't get anywhere. I've tried for a couple hours now, but still cannot progress. Maybe the refinement lemma can be used here to construct a second partition and proceed from there. I think the ultimate goal here is to relate the lower and upper Darboux sums to the integral, which I cannot seem to do. Any help is appreciated.

My attempt:

Since $$f$$ is integrable, the value of $$\int_{a}^{b} f$$ exists. Also, for any partition $$P$$, $$L(f, P) \leq \int_{a}^{b} \leq U(f, P)$$, which means that $$L(f, P_{1}) \leq U(f, P_{2})$$.

Now, there are a sequence of partitions $$\{P_{n}\}$$ for $$f$$ on $$[a, b]$$ such that $$\lim_{n\to \infty} L(f, P_{n}) = \int_{a}^{b} f$$, and there is a sequence of partitions $$\{Q_{n}\}$$ for $$g$$ on $$[a, b]$$ such that $$\lim_{n\to \infty} U(g, Q_{n}) = \int_{a}^{b} g.$$

But since $$L(f, P_{1}) \leq U(f, P_{2})$$, we have $$L(f, P_{n}) \leq U(f, Q_{n})$$ for every $$n$$. So, $$\lim_{n \to \infty} L(f, P_{n}) \leq \lim_{n\to\infty} U(f, Q_{n})$$.

Finally $$\int_{a}^{b} f = \lim_{n\to\infty} L(f, P_{n}) \leq \lim_{n\to \infty} U(f, Q_{n}) \leq \lim_{n\to\infty} U(g, Q_{n}) = \int_{a}^{b} g$$, as desired.

• Are you trying to prove that $\int f \le \int g$ or are you trying to prove the $L(f,P_1) \le U(f,P_2)$? – Alex Ortiz Oct 29 '18 at 19:02

The comparison of upper and lower Darboux sums for a given function over two partitions can be done by using the union of two partitions and you should try to prove it.

The monotone nature of the integral on the other hand is equivalent to the fact that if a function is non-negative then so is its integral. Why? Because any lower Darboux sum is non-negative and the integral is supremum of all lower sums so that it must also be non-negative. I don't think you need to compare upper and lower Darboux sums here.

Your proof is unnecessarily long. Just take any sequence $$P_n$$ of partitions such that $$\operatorname {gap} P_n\to 0$$ as $$n\to\infty$$ and then $$\int_{a} ^{b} f(x) \, dx=\lim_{n\to\infty} L(f, P_n) \leq \lim_{n\to\infty} L(g, P_n) =\int_{a} ^{b} g(x) \, dx$$ A Riemann integral (if it exists) is the limit of upper as well as lower Darboux sums and hence it is sufficient here to consider only one type of sum either upper or lower but no need to deal with both.

• i still cannot get anything. i have a solution of my own, but I think it's wrong because it doesn't use the union method described. – user400359 Oct 29 '18 at 17:24
• can u check my proof. i just updated my original post – user400359 Oct 29 '18 at 17:29
• @stackofhay42: see my last paragraph in updated answer. – Paramanand Singh Oct 30 '18 at 0:50

Consider the union $$P:=P_1\cup P_2$$ of the two partitions. Then $$P$$ is finer than both $$P_1$$ and $$P_2$$. It follows that $$L(f,P_1)\leq L(f,P)\leq U(f,P)\leq U(f,P_2)\>,$$ as desired in the first sentence of your question.

• i still cannot get anything. i have a solution of my own, but I think it's wrong because it doesn't use the union method described. – user400359 Oct 29 '18 at 17:24