# Can we say "$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\sup_{x\in [t_{k-1},t_{k}]}f(x)=U(f)$"?

Let $$f:[0,1]\rightarrow \mathbb{R}$$ be bounded and integrable function.

Let $$R_n=\frac{1}{n}\sum_{k=1}^{n}f\left ( \frac{k}{n} \right )$$.

I need to prove "$$\lim_{n\rightarrow \infty }R_n=\int_{0}^{1}f(x)dx$$".

I will write down my approach first, so you can clearly see what the question in the title means.

My approach:

Let partition $$P=\{0,\frac{1}{n},\frac{2}{n},...,1\}$$

Then note that, $$\inf_{x\in \left [ 0,\frac{1}{n} \right ]}f(x)\leq f\left ( \frac{1}{n} \right )\leq \sup_{x\in \left [ 0,\frac{1}{n} \right ]}f(x)$$ $$\inf_{x\in \left [ \frac{1}{n},\frac{2}{n} \right ]}f(x)\leq f\left ( \frac{2}{n} \right )\leq \sup_{x\in \left [ \frac{1}{n},\frac{2}{n} \right ]}f(x)$$ $$\cdot \cdot \cdot$$ $$\inf_{x\in \left [ \frac{n-1}{n},1 \right ]}f(x)\leq f\left ( \frac{n}{n} \right )=f(1)\leq \sup_{x\in \left [ \frac{n-1}{n},1 \right ]}f(x)$$

So, $$\sum_{k=1}^{n}\inf_{x\in \left [ t_{k-1},t_{k} \right ]} f(x)\leq \sum_{k=1}^{n}f\left ( \frac{k}{n} \right )\leq \sum_{k=1}^{n}\sup_{x\in \left [ t_{k-1},t_{k} \right ]}f(x)$$

$$\Rightarrow \frac{1}{n}\sum_{k=1}^{n}\inf_{x\in \left [ t_{k-1},t_{k} \right ]} f(x)\leq \frac{1}{n}\sum_{k=1}^{n}f\left ( \frac{k}{n} \right )\leq \frac{1}{n}\sum_{k=1}^{n}\sup_{x\in \left [ t_{k-1},t_{k} \right ]}f(x)\; \; \; \cdot \cdot \cdot \bigstar$$ $$\Rightarrow \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\inf_{x\in \left [ t_{k-1},t_{k} \right ]} f(x)\leq \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}f\left ( \frac{k}{n} \right )\leq \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\sup_{x\in \left [ t_{k-1},t_{k} \right ]}f(x)$$

From this step, I wanted to show that $$\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\inf_{x\in \left [ t_{k-1},t_{k} \right ]} f(x)=L(f)\; \; \; \text{and} \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\sup_{x\in \left [ t_{k-1},t_{k} \right ]}f(x)=U(f)$$

so that this could automatically implies $$\lim_{n\rightarrow \infty }R_n=\int_{0}^{1}f(x)dx$$.

However, I am not sure where to start to show them.

One idea that I have is substracting $$\frac{1}{n}\sum_{k=1}^{n}\inf_{x\in \left [ t_{k-1},t_{k} \right ]} f(x)$$ from each side in step $$\bigstar$$ because "f is integrable" implies "$$\forall \varepsilon > 0,\; \exists P$$ s.t $$U(f,P)-L(f,P)<\varepsilon$$"

Can I get some help?

Update:

As we substract $$\frac{1}{n}\sum_{k=1}^{n}\inf_{x\in \left [ t_{k-1},t_{k} \right ]} f(x)$$ from each side, then we get $$\frac{1}{n}\sum_{k=1}^{n}\left (\sup_{x \in [t_{k-1},t_k]}f(x)-\inf_{x \in [t_{k-1},t_k]}f(x) \right )=U(f,P)-L(f,P)$$ on the right hand side.

Thus, $$\left | \frac{1}{n}\sum_{k=1}^{n}f\left ( \frac{k}{n} \right ) \right |\leq U(f,P)-L(f,P)$$

We can take $$N$$ large enough to make it satisfy the following condition, $$\forall \varepsilon >0\: \: \exists N \: \: s.t\: \: |U(f,P)-L(f,P)|<\varepsilon \; \; \forall n>N$$ because there always some refinement of $$P$$ that can make difference smaller.

Therefore $$\lim [U(f,P)-L(f,P)]=0$$

And, thus

$$\lim L(f,P) = L(f) = U(f) = \lim U(f,P)$$ This gives us, $$\lim_{n\rightarrow \infty }R_n=\int_{0}^{1}f(x)dx$$

Does my updated part look right?

• The result follows by definition of Riemann integral. If a function is Riemann integrable then its Riemann sum tends to its integral. May 9, 2021 at 11:54
• It appears you didn't understand the point raised in my last comment. What is your definition of $\int_0^1 f(x) \, dx$? Under one of the definitions (given by Riemann) your result in question is nothing more than an immediate consequence of definition. May 14, 2021 at 6:02
• You seriously don't need to deal with upper and lower Darboux sums and their limits. May 14, 2021 at 6:03
• It means that you know the Darboux sums and the criterion of integrability in terms of Darboux sums. See this answer which shows that upper Darboux sums tend to $U(f)$ and lower Darboux sums tend to $L(f)$. Both these limits are equal (as $f$ is integrable) and then Riemann sum is sandwiched between these sums so it also tends to same limit. May 14, 2021 at 6:52
• Also please update your question with definition of integrability which you mention in comments. The problem or its equivalent has been amply discussed in many questions on this website. May 14, 2021 at 6:53

I think you have misunderstood the point of the question. The main point is to show that your sequence of partitions is somehow representative of any sequence of partitions which achieve $$L(f)$$ and $$U(f)$$. Let $$P_n$$ denote the partition with increment $$\frac{1}{n}$$. You want to show $$\lim_{n\to \infty}L(f,P_n)=L(f)$$ and $$\lim_{n\to \infty}U(f,P_n)=U(f)$$. You can prove these separately and the proofs are analogous.

Note the partitions $$P_n$$ are not nested and partitions containing more points do not necessarily converge to $$L(f)$$. The key issue is that the maximum size of the increments decreases with $$n$$. With this in mind, we need to show:

1. $$L(f,P_n)\leq L(f)$$

This is clear as $$L(f):=\sup\{L(f,P)\}$$ where the supremum is taken over all partitions.

1. For all $$\epsilon>0$$ there exists $$N$$ such that for all $$n\geq N$$ we have $$L(f)-\epsilon\leq L(f,P_n)$$.

First choose a partition $$Q:=\{0=q_0<... such that $$L(f)-L(f,Q)<\frac{\epsilon}{2}$$. Say $$Q$$ has $$M$$ partitions. Let $$N$$ be chosen so that $$\frac{4M\max\{|f|\}}{N}< \frac{\epsilon}{2}$$ and $$\frac{1}{N}<\min_n\{q_n-q_{n-1}\}$$. $$L(f,P_n)$$ can be calculated as the sum over intervals containing points of $$Q$$ and those that do not. Consider $$i_n\leq j_n$$ such that $$\frac{i_n}{n}\leq q_{n-1}<\frac{i_n+1}{n}$$ and $$\frac{j_n}{n}\leq q_n<\frac{j_n+1}{n}$$. Then,

$$(q_n-q_{n-1})\inf_{x\in[q_{n-1},q_n]}f(x)\leq \sum_{k=i_n}^{j_n+1}(\frac{k+1}{n}-\frac{k}{n})\inf_{x\in[q_{n-1},q_n]}f(x)\leq \sum_{k=i_n+1}^{j_n}(\frac{k+1}{n}-\frac{k}{n})\inf_{x\in[\frac{k}{n},\frac{k+1}{n}]}f(x)+\frac{2\max\{|f|\}}{n}$$

Summing over $$n$$ we get,

$$L(f,Q)=\sum_{n=1}^M (q_n-q_{n-1}) \inf_{x\in[q_{n-1},q_n]}f(x)\leq L(f,P_n)+\frac{2M\max\{|f|\}}{n}-\sum_{n=1}^M \inf_{x\in[\frac{i_n}{n},\frac{i_n+1}{n}]}f(x)+\inf_{x\in[\frac{j_n}{n},\frac{j_n+1}{n}]}f(x)\leq L(f,P_n)+\frac{\epsilon}{2}$$

Therefore $$L(f)

Putting 1) and 2) together we have that for all $$\epsilon>0$$ there exists $$N$$ such that for all $$n\geq N$$ we have $$L(f)-\epsilon\leq L(f,P_n)\leq L(f)$$ and hence $$L(f,P_n)$$ is Cauchy and $$\lim_{n\to \infty} L(f,P_n)=L(f)$$