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.

  • $\begingroup$ 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

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?


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