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I know there is a Proposition:

for $f(x)$ is bounded on $[a,b]$,then $f(x)$ is integrable if and only if given $\epsilon>0$, there exists a partition such that $U(f,P)-L(f,P)<\epsilon$

But my question is that:

Assume $f(x)$ is bounded and integrable on $[a,b]$, then given $\epsilon>0$ does it must exist a $\delta>0$ for any partition $|P|<\delta$,$U(f,P)-L(f,P)<\epsilon$.

If it is true, how to prove it? Thanks in advance.

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  • $\begingroup$ what means $|P|<\delta$? Maybe you try to note $\Delta x_j<\delta$? $\endgroup$ – Masacroso Jun 21 '16 at 7:18
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    $\begingroup$ $|P|$ means the mesh of any partition of $[a,b]$ $\endgroup$ – Spaceship222 Jun 21 '16 at 7:20
  • $\begingroup$ what you mention is true and not difficult to prove but yes it is not an obvious and trivial result. see this answer math.stackexchange.com/a/1255344/72031 Especially see the proposition (3) of that answer with its proof. $\endgroup$ – Paramanand Singh Jun 21 '16 at 9:10
  • $\begingroup$ BTW in case it is not clear if $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}$ is a partition of $[a, b]$ then $|P| = \max_{k = 1}^{n}(x_{k} - x_{k - 1})$ so that $|P|$ is the length of largest sub-interval created by points of partition $P$. $\endgroup$ – Paramanand Singh Jun 21 '16 at 9:14
  • $\begingroup$ @ParamanandSingh I don't know how to construct $S_1$ and $S_2$ in your answer before,can you give me more detail about it? $\endgroup$ – Spaceship222 Jun 21 '16 at 10:11
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I will provide a proof beginning with some explanatory notes. Most of what follows is taken from Tom Apostol's Mathematical Analysis (it is one of the very good books and appears in my bibliography). The content in the beginning of this answer is repetitive (for those who are well versed) and the reader may jump directly after the first fold (starting with "Integrability condition 2") if he prefers to do so.

Apostol defines Riemann integral as follows:

Let $[a, b]$ be a closed interval. A partition $P$ of $[a, b]$ is a set of points $$P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$$ such that $$a = x_{0} < x_{1} < x_{2} < \cdots < x_{n} = b$$ A partition $P'$ is said to be finer than a partition $P$ if $P' \supseteq P$.

Definition 1: Let $f$ be a function bounded on $[a, b]$ and let $P = \{x_{0}, x_{1}, \ldots, x_{n}\}$ be a partition of $[a, b]$. A sum of the form $$S(P, f) = \sum_{k = 1}^{n}f(t_{k})(x_{k} - x_{k - 1})$$ is called a Riemann sum for $f$ over partition $P$ where $t_{k}$ is any point in the interval $[x_{k - 1}, x_{k}]$. The function $f$ is said to be Riemann integrable over $[a, b]$ if there is a real number $I$ such that for every $\epsilon > 0$ there is a partition $P_{\epsilon}$ of $[a, b]$ which has the following property: $$|S(P, f) - I| < \epsilon$$ whenever $P$ is a partition of $[a, b]$ finer than $P_{\epsilon}$. The number $I$ is then said to be the Riemann integral of $f$ over $[a, b]$ and we write $$I = \int_{a}^{b}f(x)\,dx$$

The Riemann integral is defined as a (complicated) limit of Riemann sums when the partitions are made finer and finer. Most other books on analysis also define Riemann integral as limit of Riemann sums but the limit is not taken by making partitions finer and finer. Rather they use the concept of mesh or norm (this term is used by Apostol) of a partition. If $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ is a partition of $[a, b]$ and then norm $||P||$ is defined by $||P|| = \max_{k = 1}^{n}(x_{k} - x_{k - 1})$ so that norm of a partition $P$ is the length of largest sub-interval of $[a, b]$ created by points of the partition $P$. The definition of Riemann integral based on norm is as follows:

Definition 2: Let $f$ be bounded on $[a, b]$. Then $f$ is said to be Riemann integrable over $[a, b]$ if there is a number $I$ such that for every $\epsilon > 0$ there is a number $\delta > 0$ with the following property: $$|S(P, f) - I| < \epsilon$$ for all partitions $P$ of $[a, b]$ with $||P|| < \delta$. The number $I$ is then said to be the Riemann integral of $f$ over $[a, b]$ and we write $$I = \int_{a}^{b}f(x)\,dx$$

Thus in this definition the limit is taken not by necessarily making finer and finer partitions, but by using partitions with smaller and smaller norms. When we make a partition finer by adding additional points to it, then the norm decreases (or stays same) so making a partition finer in general leads to smaller norm, but making norm smaller does not necessarily correspond to making partitions finer.

It is then a remarkable fact that the above two definitions of Riemann integrability are equivalent and the proof of this fact is not trivial. The definition based on "finer" partitions is easier to work with and hence Apostol prefers it compared to another definition. Next it turns out that Riemann sums are also somewhat difficult to deal with because of the arbitrary nature of points $t_{k}$ involved. Hence it is important to make a transition to Darboux sums (and these are the topic of the current question).

Darboux sums replace $f(t_{k})$ with supremum (Upper Darboux sum) and infimum (lower Darboux sum) of $f$ on interval $[x_{k - 1}, x_{k}]$ and we define $$L(P, f) = \sum_{k = 1}^{n}m_{k}(x_{k - 1} - x_{k}), U(P, f) = \sum_{k = 1}^{n}M_{k}(x_{k - 1} - x_{k})$$ where $$M_{k} = \sup\,\{f(x)\mid x \in [x_{k - 1}, x_{k}]\},\, m_{k} = \inf\,\{f(x)\mid x \in [x_{k - 1}, x_{k}]\}$$

Note that lower Darboux sum increases (or stays same) when partition is made finer and upper Darboux sum decreases (or stays same) when partition is made finer. Further any lower Darboux sum does not exceed any upper Darboux sum. Since $M_{k}, m_{k}$ are supremum and infimum on $f$ on $[x_{k - 1}, x_{k}]$ it is possible to find points $t_{k}, t_{k}'$ in $[x_{k - 1}, x_{k}]$ such that $f(t_{k})$ is near $M_{k}$ and $f(t_{k}')$ is near $m_{k}$ and hence it is possible to find Riemann sums near both $L(P, f)$ and $U(P, f)$. Moreover any Riemann sum is clearly sandwiched between lower and upper Darboux sums. Using these facts and not so difficult arguments it is possible to prove the following (keeping in mind the definition 1 of Riemann integrability):

Integrability Condition 1 A bounded function $f$ is Riemann integrable over $[a, b]$ in the sense of definition 1 if and only if for every $\epsilon > 0$ it is possible to find a partition $P_{\epsilon}$ of $[a, b]$ such that $U(P_{\epsilon}, f) - L(P_{\epsilon}, f) < \epsilon$.


Now we show that the above condition of integrability is equivalent to the following (and this is the question asked here):

Integrability Condition 2 A function $f$ bounded on $[a, b]$ is Riemann integrable over $[a, b]$ if and only if for every $\epsilon > 0$ there is a number $\delta > 0$ such that $$U(P, f) - L(P, f) < \epsilon$$ whenever norm $||P|| < \delta$.

Note that if we satisfy this condition 2 then we trivially satisfy condition 1. Proving the converse is difficult. Let us suppose that $f$ meets the condition 1 mentioned above. Then for every $\epsilon > 0$ there is a partition $P_{\epsilon}$ of $[a, b]$ such that $$U(P_{\epsilon}, f) - L(P_{\epsilon}, f) < \frac{\epsilon}{2}$$ Let $N$ be the number of sub-intervals created by $P_{\epsilon}$ so that we can assume $$P_{\epsilon} = \{x_{0}', x_{1}', \ldots, x_{N}'\}$$ Also let $$K = \sup\,\{|f(x)|,\, x \in [a, b]\} + 1$$ and let $\delta = \epsilon / (4NK) > 0$.

If $$P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$$ is any partition of $[a, b]$ with norm $||P|| < \delta$ then we divide sum $$S = U(P, f) - L(P, f) = \sum_{k = 1}^{n}(M_{k} - m_{k})(x_{k} - x_{k - 1})$$ into two sums $S = S_{1} + S_{2}$ where $S_{1}$ has terms corresponding to the index $k$ such that $[x_{k - 1}, x_{k}]$ contains no points of $P_{\epsilon}$ and $S_{2}$ consists of remaining terms. Thus for the index $k$ used in making sum $S_{1}$ it is ensured that interval $[x_{k - 1}, x_{k}]$ is contained inside one of intervals $[x_{j - 1}', x_{j}']$ made by partition $P_{\epsilon}$. Hence $$S_{1} \leq U(P_{\epsilon}, f) - L(P_{\epsilon}, f) < \frac{\epsilon}{2}$$ For $S_{2}$ we see that intervals which make $S_{2}$ contain points of $P_{\epsilon}$ and hence the number of intervals is at max $N$ and the $M_{k} - m_{k}$ is at max $2K$ so $$S_{2} < N \cdot 2K \cdot \delta < \frac{\epsilon}{2}$$ Together we see that $$S = U(P, f) - L(P, f) = \sum_{k = 1}^{n}(M_{k} - m_{k})(x_{k} - x_{k - 1}) = S_{1} + S_{2} < \epsilon$$ for any $P$ with $||P|| < \delta$ and hence the condition 2 for integrability is also satisfied.


Since it is easy to map from Darboux sums to Riemann sums (as explained earlier) it follows from the equivalence of conditions 1 and 2 quoted above that the two definitions of Riemann integrability are equivalent.

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    $\begingroup$ Thanks.I just now understood your proof in your answer before.Nice proof! $\endgroup$ – Spaceship222 Jun 21 '16 at 10:29
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    $\begingroup$ @Spaceship222: Thanks to you too! The fact you have asked in your question is mostly overlooked by many students and professors and textbooks alike (thus Rudin does not mention this in his textbook) and its great that you raised this point. $\endgroup$ – Paramanand Singh Jun 21 '16 at 10:32
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I will try to answer with slightly different approach. It almost follows paramanand's approach.


Let $f$ be a bounded function on $[a,b]$. The set $P=\{a=x_0<x_1...<x_i<....<x_n=b\}$ is defined to be a partition of $[a,b]$. If we additionally specify a set $T=\{t_i\}_1^n, t_i \epsilon[x_{i-1},x_i]$, $(P,T)$ will be referred as a tagged partition with tags in T.

Let $U(f,P)=\sum_i^nM_i(x_i-x_{i-1})$ where $M_i=sup\{f(x)\}_{[x_{i-1},x_i]}$
Let $K=sup\{|f(x)|\}_{[a,b]}$
$L(f,P)=\sum_i^nm_i(x_i-x_{i-1})$ where $m_i=inf\{f(x)\}_{[a,b]}$

As $f$ is bounded, $inf\{U(f,P)\}=\int^-f$ and $sup\{L(f,P)\}=\int_-f$ both exist.

$\forall \epsilon>0 \exists P, U(f,P)-\int^-f<\epsilon$

$P \subset P^{'} \implies U(f,P^{'})\le U(f,P)$

Let $Q$ be any partition and let $P^{'}=P \bigcup Q\implies P\subset P^{'}\implies \int^-f \le U(f,P^{'})\le U(f,P) < \int^-f+\epsilon$

Let $Q=\{a=x_0<x_1...<x_j<....<x_m=b\}$
$U(f,Q)-U(f,P^{'})=\sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[M_i(x_i-x_{i-1})-M_j^{'}(x_j-x_{i-1})-M_j^{"}(x_i-x_j)]$

$|U(f,Q)-U(f,P^{'})|=U(f,Q)-U(f,P^{'})$

$=|\sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[M_i(x_i-x_{i-1})-M_j^{'}(x_j-x_{i-1})-M_j^{"}(x_i-x_j)]|$

$\le \sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[|M_i|(x_i-x_{i-1})+|M_j^{'}|(x_j-x_{i-1})+|M_j^{"}|(x_i-x_j)]$

$\le \sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[2K(x_i-x_{i-1})]$

Choose $Q$ with $||Q||< \epsilon / (2nK)$ we get

$U(f,Q)-U(f,P^{'}) < \epsilon$ we already have

$U(f,P^{'})\le U(f,P) < \int^-f+\epsilon$

so $U(f,Q)-\int^-f<2\epsilon$

i.e $\forall \epsilon > 0 \exists \delta_1 > 0,||Q|| < \delta_1 \implies U(f,Q)-\int^-f<\epsilon$

similarly we obtain

$\forall \epsilon > 0 \exists \delta_2 > 0, ||Q|| < \delta_2 \implies\int_-f - L(f,Q)<\epsilon$

In case $f$ is darboux integrable, $\int_-f = \int^-f =I$

Taking $\delta < min\{\delta_1, \delta_2 \}$ and combining last two inequalities

$\forall \epsilon > 0 \exists \delta > 0, ||Q|| < \delta \implies U(f,Q)-L(f,Q) < 2\epsilon$

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