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?


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?

  • 1
    $\begingroup$ The result follows by definition of Riemann integral. If a function is Riemann integrable then its Riemann sum tends to its integral. $\endgroup$
    – Paramanand Singh
    May 9, 2021 at 11:54
  • 1
    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    May 14, 2021 at 6:02
  • 1
    $\begingroup$ You seriously don't need to deal with upper and lower Darboux sums and their limits. $\endgroup$
    – Paramanand Singh
    May 14, 2021 at 6:03
  • 1
    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    May 14, 2021 at 6:52
  • 2
    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    May 14, 2021 at 6:53

1 Answer 1


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 $Q:=\{0=q_0<...<q_{M}=1\}$ 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 $$L(f)<L(f,Q)+\frac{\epsilon}{2}\leq L(f,P_n)+\epsilon $$

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 $\lim_{n\to \infty} L(f,P_n)=L(f)$


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