Why are these definitions equivalent for $\alpha=1$?

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I am getting confused over what I can only assume is a technicality of a limit in the following definitions;

Definition 1:

Let $I=(a,b)\subset \mathbb{R}$ be an interval and $f, F:I\rightarrow \mathbb{R}$ be functions. We say that $F$ is an indefinite MC integral of $f$ if there exists a strictly increasing function $\varphi :I \rightarrow \mathbb{R}$ such that;

$ \displaystyle\lim_{y\to x}\frac{F(y)-F(x)-f(x)(y-x)}{\varphi(y)-\varphi(x)}=0,\,\,x\in I\,\,\,(1)$

Now the second one requires some notation;

Write $I(x,r)=(x-r,x+r)$. If $\alpha>0$ and $I=I(x,r)$ then $\alpha I = I(x,\alpha r)$. Write $\Delta_I F=F(b)-F(a)$ if $I=(a,b)$.

Definition 2;

We say that $F$ is an indefinite $MC_\alpha$-integral of $f:J\rightarrow \mathbb{R}$ if there exists a strictly increasing function $\varphi :J\rightarrow \mathbb{R}$ such that (letting $I=I(x,r)$);

$ \displaystyle\lim_{r\to 0}\frac{\Delta_I F - f(x)|I|}{\Delta_{\alpha I} \varphi}=0,\,\,x\in J\,\,\, (2)$

With $\alpha = 1$ we recover the MC-integral i.e definition 1.

It is this last sentence I don't understand, with $\alpha = 1$ we get

$ \displaystyle\lim_{r\to 0}\frac{F(x+r)-F(x-r) - f(x)(2r)}{\varphi(x+r)-\varphi(x-r)}=0,\,\,x\in I$

and by setting $y=x+r$ in (1) we get

$ \displaystyle\lim_{r\to 0}\frac{F(x+r)-F(x) - f(x)(r)}{\varphi(x+r)-\varphi(x)}=0,\,\,x\in I$

And I'm not really sure why these are equivalent there is a fixed $x$ in one and not in the other and we know nothing about continuity etc of the functions.

I would really appreciate if someone could help.