$\mu_{n-1}$ is the $n-1$ dimensional measure and $\operatorname{meas}$ is the $n$-dimensional one.

$I(\varrho)$ is the ball of radius $\varrho$ around a fixed point $y$ in the domain $\Omega\subset \mathbb{R}^n$ and $A(t,\varrho)$is the subset of $I(\varrho)$ where a certain function $u$ is greater than $t$.

Surely $I(\varrho) \geq 2\tau(t;\varrho)$ infact, $$\operatorname{meas}{A(t;\varrho)}+\operatorname{meas}(I(\varrho)\setminus A(t;\varrho))= \operatorname{meas}I(\varrho)$$

**It seems to me that it is claimed $x > y $ implies $ A-x > A-y$**

Fortunately, it turns out the inequality still holds but you need a bit of work (EDIT)

**Original article: De Giorgi's original article
pages 5-6**

**EDIT:**

Let us assume that $\operatorname{meas}(I(\varrho))=1$ then $\tau(t;\varrho)=x\leq \frac{1}{2}$ what we need to prove is that for $\alpha < 1$ $$(1-x)^{\alpha} + x^{\alpha} -1 \geq 2x^{\alpha}-(2x)^{\alpha}$$

but considering $$f(x)=\left( \frac{1-x}{x}\right) ^{\alpha} -\frac{1}{x^{\alpha}}$$

we can prove that it is decreasing in $(0,\frac{1}{2})$ and $f(\frac{1}{2})=1-2^{\alpha}$

From this we obtain the inequality and hence justify the line.

Due to the several mistakes the question has changed a lot since I posted it, just to be clear I am asking the following now:

**1) Is my "fixing" of the proof alright or am I oversimplifying things?**

**2) Do you think it was an actual mistake that fortunately did not lead to a problem or it is something obvious enough to be omitted in an article?**

P.S. sorry if the question is not specific enough, I think it was when I first posted it!