3
$\begingroup$

Let $f[a,b]\to\mathbb{R}$ be an integrable function. Prove the following, using only the definition of the integral $$\text{For any}~c>0,\int^b_af(x)dx=c\int^{b/c}_{a/c}f(cx)dx$$ Hint: A careful choice of notation is essential in solving this problem, you should consistently write $P$ to denote a partition of $[a,b]$ and $P'$ a partition of $[a/c,b/c].$ You may want to choose $P$ and $P'$ to be related in some way. With this notation, you can also write $m_j,M_j$ to refer to the inf and sup of $f(x)$ for $x$ in the $j$th interval of $P$, and $m_j',M_j'$ for the inf and sup of $f(cx)$ in the $j$th interval of $P'$.

$($The question is from this online note$)$

This is a short summary of the integral definition

$\def\box#1#2{\boxed{\underline{\text{#1}}\\#2}} \def\verts#1{\left\vert#1\right\vert}$ $\box{Def. Integrable Function Single Variable} {\text{A function $f:[a,b]\to\mathbb{R}$ is integrable if it is bounded and $\underline{I^b_a}f=\overline{I^b_a}f.$ When this hold, we define}\\ \int_a^bf(x)dx=\underline{I^b_a}f=\overline{I^b_a}f, \text{ the integral of $f$ over $[a,b]$.}}$

Here $\underline{I^b_a}f=\sup_PL_Pf$, and $\overline{I^b_a}f=\inf_PU_Pf$

where $P$ is a partition of $[a,b]$, that $L_P f=\sum_{j=1}^Jm_j\text{length}(I_j)$ and $U_Pf=\sum_{j=1}^JM_j\text{length}(I_j)$

and $m_j=\inf\{f(x):x\in I_j\}\hspace{5ex}M_j=\sup\{f(x):x\in I_j\}$


My thought

Based on my understanding, the definition can be written as \begin{align} \int_a^bf(x)dx=&\sup\left\{\sum_{i=1}^{\verts{P}-1}\left[\inf_{x\in\left[x_i,x_{i+1}~~\right]}f(x)\right](x_{i+1}-x_i):\text{$P$ is a partition of $[a,b]$}\right\}\\ =&\inf\left\{\sum_{i=1}^{\verts{P}-1}\left[\sup_{x\in\left[x_i,x_{i+1}~~\right]}f(x)\right](x_{i+1}-x_i):\text{$P$ is a partition of $[a,b]$}\right\}\\ c\int_{a/c}^{b/c}f(x)dx=&\sup\left\{\sum_{i=1}^{\verts{P'}-1}\left[\inf_{x\in\left[x_i,x_{i+1}~~\right]}f(cx)\right](x_{i+1}-x_i):\text{$P'$ is a partition of $\left[\frac{a}{c},\frac{b}{c}\right]$}\right\}\\ =&\inf\left\{\sum_{i=1}^{\verts{P'}-1}\left[\sup_{x\in\left[x_i,x_{i+1}~~\right]}f(cx)\right](x_{i+1}-x_i):\text{$P'$ is a partition of $\left[\frac{a}{c},\frac{b}{c}\right]$}\right\} \end{align}

However, I still can't see how to write this proof, could someone help me.

$\endgroup$
2
  • $\begingroup$ Can you start with the second integral and represent it as a Riemann sum and go from there? $\endgroup$
    – Henry Lee
    Jul 19 '20 at 23:06
  • 1
    $\begingroup$ @HenryLee I think it's ok to start with the second integral, but the definition is using Darboux sums $\endgroup$
    – Manx
    Jul 19 '20 at 23:15
2
$\begingroup$

Consider the partition of $[a,b]$ as $P=\{a=x_0,x_1,...,x_{n-1},x_n=b\}$

Hence partition of $[a/c,b/c]= \{a/c=x_0/c,x_1/c,...,x_{n-1}/c,x_n/c=b/c\}$
Let $M_j=\sup \{f(s): x_{j-1}\le s\le x_j\}, m_j=\inf \{f(s): x_{j-1}\le s\le x_j \}$

Let $M_j'=\sup \{f(cs): x_{j-1}/c\le s\le x_j/c\}, m_j'=\inf\{f(cs): x_{j-1}/c\le s\le x_j/c\}$

Do you see why $M_j=M_j'$ and $m_j=m_j'$?

Upper sum (Darboux's upper sum) for $f(t) $ over $P=\sum_{j=1}^{n}M_j(x_j-x_{j-1})=\sum_{j=1}^{n}cM_j' (x_j/c-x_{j-1}/c)$, where $\sum_{j=1}^{n}M_j' (x_j/c-x_{j-1}/c)$ is upper sum of $f(ct)$ over $[a/c,b/c]$ etc.

Can you take it from here?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.