# Show $\forall c>0$ have $\int^b_af(x)dx=c\int^{b/c}_{a/c}f(cx)dx$ from the definition of integral

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.

• Can you start with the second integral and represent it as a Riemann sum and go from there? Jul 19 '20 at 23:06
• @HenryLee I think it's ok to start with the second integral, but the definition is using Darboux sums
– 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.