# Proving that a linear transform from a domain to the domain of another function (f) can be composed with f

... to give a function with the same properties as f.

Suppose $f:[a,b]\rightarrow\mathbb{R}$

Suppose $L:[c,d]\rightarrow[a,b]$ and that this is a linear transform (a bijection)

I want to pove that $g = f\circ L:[c,d]\rightarrow\mathbb{R}$ and that g has all the properties of $f$

Context (why I want to show this)

There is a homework question involving step functions, show that for a step function $a:[a,b]\rightarrow\mathbb{R}$ that for a fixed $k>0 \in \mathbb{R}$ that $b:[ka,kb]\rightarrow\mathbb{R}$ by $b=a(\frac{x}{k})$ that $b$ is a step function.

It actually says prove.

Then prove that $\int^{kb}_{ka}b=k\int^b_aa$

I want to just use a substitution to show this, but we are doing this to lead to the analysis side of integration. Regardless linear algebra can give me a proof without actually caring about the nature of the functions a and b.

So yes, I want to prove the first thing. It is intuitively obvious.

How I've shown this

Just by using composition, it's like 3 lines, I define $L(x)=\frac{x}{k}$ and that $L:[ka,kb]\rightarrow[a,b]$, then just compose that.

How can I make this more formal. I am using the idea that a bijection is just a mapping between two sets, and by using L to take a set to the domain of the function I want to compose it with, and that is is a bijection, means that I can map the domain of L to the target of whatever I want to compose it with.

-
Suppose you have a step function $s:[a,b]\to\Bbb R$ with partition $P=\{a=x_0,\ldots,x_{n-1},x_n=b\}$ which has the value $s_i$ at the $i$-th interval $(x_i,x_{i+1})$. The value at end points can be arbitrary. It is immaterial to this discussion.
Let $k>0$. Define $t:[ka;kb]\to\Bbb R$ by $t(x)=s\left(\dfrac{x}{k}\right)$. Then $t$ is also a step function, with partition $P'=\{ka=kx_0,kx_1,\ldots,kx_{n-1},kx_n=kb\}$ and has the same values $t_i=s_i$.
By definition $$\int_a^b s=\sum_{i=0}^{n-1} s_i(x_{i+1}-x_i)$$ Thus $$\int_{ka}^{kb} t=\sum_{i=0}^{n-1} t_i(kx_{i+1}-kx_i)=k\sum_{i=0}^{n-1} s_i(x_{i+1}-x_i)=k\int_a^b s$$