Riemann Integral: Proving $\int_a^b f = \int_a^c f + \int_c^b f$

[evinda]

Hello again! (Blush)

Let $f:[a,b] \to \mathbb{R}$ bounded and $c \in (a,b)$.Then $f$ is integrable at $[a,b]$ iff $f$ is integrable at $[a,c]$ and $[c,b]$.In this case,we have $\int_a^b f = \int_a^c f + \int_c^b f$.
The proof for the direction $\Rightarrow$ is like that:
Suppose that $f$ is integral.Let $\epsilon>0$.As $f$ is integrable there is a partition $P=\{a=t_0<t_1<...<t_n=b\}$ of $[a,b]$ such that $U(f,P)-L(f,P)< \epsilon$.
The point $c$ is in an interval $[t_i,t_{i+1}]$.Let's suppose,without loss of generality,that $c$ is not a endpoint of $[t_i,t_{i+1}]$.
So,we have the partitions:
$P_{1}=\{a=t_{0}<...<t_i<c\}$ of $[a,c]$ and $P_2={c<t_{i+1}<...<t_{n}=b}$ of $[c,b]$.
Therefore,$U(f,P_1)-L(f,P_1)\overset{1}{=}$$[U(f,P \cup \{c\})-L(f,P \cup \{c\})]-U(f,P_2)-L(f,P_2)$$ \leq U(f,P \cup \{c\})-L(f,P \cup \{c\}) \leq U(f,P)-L(f,P) < \epsilon$

But..how do we get to $\overset{1}{=}..$?? Isn't it $P=P_1 \cup P_2 \cup \{c\}$ ?

 

[Amer]

Hello again! (Blush)

Let $f:[a,b] \to \mathbb{R}$ bounded and $c \in (a,b)$.Then $f$ is integrable at $[a,b]$ iff $f$ is integrable at $[a,c]$ and $[c,b]$.In this case,we have $\int_a^b f = \int_a^c f + \int_c^b f$.
The proof for the direction $\Rightarrow$ is like that:
Suppose that $f$ is integral.Let $\epsilon>0$.As $f$ is integrable there is a partition $P=\{a=t_0<t_1<...<t_n=b\}$ of $[a,b]$ such that $U(f,P)-L(f,P)< \epsilon$.
The point $c$ is in an interval $[t_i,t_{i+1}]$.Let's suppose,without loss of generality,that $c$ is not a endpoint of $[t_i,t_{i+1}]$.
So,we have the partitions:
$P_{1}=\{a=t_{0}<...<t_i<c\}$ of $[a,c]$ and $P_2={c<t_{i+1}<...<t_{n}=b}$ of $[c,b]$.
Therefore,$U(f,P_1)-L(f,P_1)\overset{1}{=}$$[U(f,P \cup \{c\})-L(f,P \cup \{c\})]-U(f,P_2)-L(f,P_2)$$ \leq U(f,P \cup \{c\})-L(f,P \cup \{c\}) \leq U(f,P)-L(f,P) < \epsilon$

But..how do we get to $\overset{1}{=}..$?? Isn't it $P=P_1 \cup P_2 \cup \{c\}$ ?

You answer yourself

without loss of generality,that $c$ is not a endpoint of $[t_i,t_{i+1}]$.

since it is not an end point for an interval like that so it is not in the partition
it is like we have a partition
x0--x1--x2--x3 : P
add c somewhere
x0--x1--c--x2--x3 : P U C

$U(P_1) = (x_1-x_0)M_1 + (c-x_1)M_2 , L(P_1) = (x_1-x_0)m_1 + (c-x_1)m_2 $

$U(P_2) = (x_2-c)M_3 + (x_3-x_2)M_4 , L(P_2) = (x_2 -c)m_3 + (x_3-x_2)m_4 $

$U(P \cup C) =(x_1-x_0)M_1 + (c-x_1)M_2 + (x_2-c)M_3 + (x_3-x_2)M_4 $
But
$U(P) = (x_1-x_0)M_1 + (x_2-x_1)M_* + (x_3-x_2)M_4 $
so
$ U(P \cup C) \leq U(P) $
M stands for the maximum of f(x) at the interval and m for the minimum of f(x) at the interval
and
$L(P) \leq L(P \cup C) $
So
$ U(P \cup C) - L(P\cup C) \leq U(P) - L(P) $...(**)
But still I can't see why :
$U(f,P_1)-L(f,P_1)\overset{1}{=}$$[U(f,P \cup \{c\})-L(f,P \cup \{c\})]-U(f,P_2)-L(f,P_2)$
I think it should be
$ U(P_1 ) - L(P_1) = U(P\cup C) - U(P_2) - ( L(P\cup C) - L(P_2)) $

My proof
$ U(P\cup C) - L(P\cup C) = U(P_1 ) - L(P_1) + U(P_2) - L(P_2) > U(P_1) - L(P_1) $
with (**) we get what we want

 

[evinda]

I haven't understood it.Does $c$ belong in the partitions $P_1$ and $P_2$ ?? (Thinking)

 

[stainburg]

I haven't understood it.Does $c$ belong in the partitions $P_1$ and $P_2$ ?? (Thinking)

Since \(\displaystyle f\in \mathcal{R}(E)\), there exists a partition \(\displaystyle P_{E}=\{t_0, t_1, ..., t_n\}\), (\(\displaystyle a=t_0\), \(\displaystyle b=t_n\)), such that \(\displaystyle U(f,P_E)-L(f,P_E)<\epsilon\). Clearly \(\displaystyle c\) must be contained in some interval \(\displaystyle [t_i,t_{i+1}]\) (for \(\displaystyle 0\leq i \leq n-1\)). Then a refinement can be put into \(\displaystyle P^{*}_{E}=\{t_0, ..., t_i, c, t_{i+1}, ..., t_n\}\). We have \(\displaystyle U(f,P^*_E)-L(f,P^*_E)<\epsilon\). Take \(\displaystyle P_{[a,c]}=\{t_0, ..., t_i, c\}\) and \(\displaystyle P_{[c,b]}=\{c, t_{i+1}, ..., t_n\}\). Hence, \(\displaystyle U(f,P^*_E)-L(f,P^*_E)=(U(f,P_{[a,c]})-L(f,P_{[a,c]}))+(U(f,P_{[c,b]})-L(f,P_{[c,b]}))<\epsilon\). Hence \(\displaystyle U(f,P_{[a,c]})-L(f,P_{[a,c]})<\epsilon\) and \(\displaystyle (U(f,P_{[c,b]})-L(f,P_{[c,b]}))<\epsilon\). Hence \(\displaystyle f\in \mathcal{R}([a,c])\), \(\displaystyle f\in \mathcal{R}([c,b])\) and \(\displaystyle \int_a^b f=\int_a^c f+\int_c^b f\).

Conversely, ...

 

[blue_raver22]

Hi there! The reason for $\overset{1}{=}$ is because we have the partition $P$ for $[a,b]$ and by adding the point $c$ to it, we are essentially creating two new partitions $P_1$ for $[a,c]$ and $P_2$ for $[c,b]$. This means that the upper and lower sums for $P$ will be equal to the upper and lower sums for $P_1$ and $P_2$ combined. This is why we can say that $U(f,P_1)-L(f,P_1)=[U(f,P)-U(f,P_2)]-[L(f,P)-L(f,P_2)]$. Hope this helps clarify!