# When does Rudin's change of variables formula break?

Here is Rudin's change of variables Theorem: My question is this: what are examples where $\varphi$ not being strictly increasing or not mapping the interval $[a,b]$ to $[A,B]$ breaks the theorem? For example, I am considering $\varphi$ a parabola with minimum point in the unit interval. Using Rudin's special case remark ($\alpha(x)=x$ and $\beta(x)=\varphi$) I keep computing $$\int_0^1 f(x)dx=\int_0^1f(\varphi(y))\varphi'(y)dy$$ with various choices of $f$, and I keep getting these integrals are equal. Does this mean $\varphi$ doesn't need such strict conditions?

If we let $a$ be the minimum point this seems surprising since $\varphi$ is only strictly increasing on $[a,1]$.

If you don't mind loosening up the hypotheses a bit:

Theorem: Suppose $\phi:[a,b] \rightarrow I$, where $I$ is an interval, is a $C^1$ function and $f$ is a continuous function on $I$. Then

$$\int_{[\phi(a),\phi(b)]} f=\int_{[a,b]} f \circ \phi \cdot \phi',$$

where $\int_{[\phi(a),\phi(b)]} f:=-\int_{[\phi(b),\phi(a)]} f$ if $\phi(b) < \phi(a)$.

Proof: Since $f$ is continuous, pick a function $F$ such that $F'=f$ (this follows from FTC). Define $H:=F \circ \phi$. We then have that $H'=f\circ \phi \cdot \phi'.$ Therefore,

$$\int_{[a,b]} f \circ \phi \cdot \phi'=\int_{[a,b]} H'=H(b)-H(a)$$

$$=F \circ \phi(b)- F \circ \phi(a)=F(\phi(b))-F(\phi(a))=\int_{[\phi(a),\phi(b)]} f.$$

$\blacksquare$

Note that we did not need $\phi$ to be increasing.