Charcteristic function not in a fractional Sobolev space

So with your method, the difficulty is the fact that you have no knowledge of the set $E$ at which the singularity occurs. However, the $H^{1/2}$ seminorm (the quantity you are trying to compute, that I will denote $\|\cdot\|_{\dot{H}^{1/2}}$) decreases when one takes a symmetric decreasing rearrangement (see e.g. Lemma 7.17 in the book Analysis by Lieb & Loss). Therefore, taking the ball $B$ centered in $0$ with the same measure as $E$, we have that $$ \|\chi_E\|_{\dot{H}^{1/2}} \geq \|\chi_B\|_{\dot{H}^{1/2}} $$ From there, one way could be to use the Fourier transform definition of $H^{1/2}$ and the exact Fourier transform of $\chi_B$ (see e.g. Fourier transform of the indicator of the unit ball). This gives us $$ \|\chi_B\|_{\dot{H}^{1/2}} = \int_{\mathbb{R}^n} |J_{n/2}(|x|)|^2 \,|x|^{1-n}\,\mathrm{d}x = C_d\int_0^\infty |J_{n/2}(r)|^2\,\mathrm{d}x $$ which is infinite since (see https://en.wikipedia.org/wiki/Bessel_function) $$ J_{n/2}(r) = (\tfrac{2}{πr})^{1/2} \cos(r-\tfrac{(n+1)\pi}{4}) + O_{r\to\infty}(\tfrac{1}{r}) $$ Therefore your integral (i.e. the seminorm) is infinite and thus $\chi_E$ is not in $H^{1/2}$.

Remark: If you want to use your computation, from my first equation you can now restrict to a ball and you have $$ \|\chi_B\|_{\dot{H}^{1/2}} = 2\int_{B} \int_{B^c} \frac{1}{|x-y |^{n+1}} \,\mathrm{d}x \,\mathrm{d}y $$ which might be easier to estimate. I suppose one can restrict this integral on a neighborhood of a point of the sphere and then say that the ball is flat near this point to also get estimates on why this integral is infinite?

This nice question led me to some thoughts, which I am going to write down in this answer.

NOTATION. The letter $C$ will always denote an irrelevant positive constant, whose value can change from line to line.

We are concerned here with the Gagliardo seminorms, which I will denote by $\dot{H}^s(\mathbb R^d)$ in compatibility with LL3.14’s answer, and which can be defined in two equivalent ways (see also this answer); $$ \lVert f\rVert_{\dot{H}^s}^2:=\int_{\mathbb R^d} \lvert \xi \rvert^{2s}\lvert \hat{f}(\xi)\rvert^2\, d\xi =C\iint_{\mathbb R^d\times \mathbb R^d} \frac{\lvert f(x+y)-f(x) \rvert^2}{\lvert y \rvert^{d+2s}}\, dxdy.$$ Here $\hat{f}$ denotes the Fourier transform. We let $$ \omega_f(y):=\int_{\mathbb R^d} \lvert f(x+y)-f(x)\rvert^2\, dx,\quad\text{ so that }\quad \lVert f \rVert_{\dot{H}^s}^2=C \int_{\mathbb R^d}\frac{\omega_f(y)}{ \lvert y \rvert^{d+2s}}\, dy. $$ The heuristic emerging from these formulations is that, for $f\in L^2$, the seminorm $\lVert f\rVert_{\dot{H}^s}^2$ is finite if and only if $\omega_f$ decays sufficiently fast at $0$, which happens if and only if $\hat{f}$ decays sufficiently fast at infinity.

In what follows, we will consider $f=\chi_E$ for some $E\subset \mathbb R^d$. In particular, we let $B$ denote the unit ball. In the nice answer of LL 3.14, the seminorm $\lVert \chi_B\rVert_{\dot{H}^2}$ is studied via the explicit decay rate of $\hat{\chi_B}$. Here we will perform the same analysis, but studying $\omega_{\chi_B}$ instead. This approach is perhaps more adherent to the OP’s initial thoughts.

Expanding the square, we see that $$\tag{1}\omega_{\chi_B}(y)=2\lvert B\rvert -2\lvert B\cap (B-y)\rvert,$$ so we are reduced to study the measure of the intersection $B\cap (B-y)$. As the following crude picture shows,

such intersection is made of two equal spherical caps. Writing the volume of such caps as an integral, we obtain $$ \lvert B\cap (B-y)\rvert = 2\lvert B^{d-1}\rvert \int_{\lvert y \rvert /2}^1 (1-z^2)^{\frac{d-1}{2}}\, dz.$$ Here $\lvert B^{d-1}\rvert $ denotes the volume of the $d-1$ dimensional ball, but it is not relevant for what follows. Indeed, we do not need an exact expression of $\lvert B\cap B-y\rvert$; an approximation to first order at $y\to 0$ will suffice. To compute such approximation, we note that $$\lvert B\cap (B-y)\rvert \Big\rvert_{y=0}=\lvert B\rvert,$$ and it is clear from the integral that $\nabla_y \lvert B\cap (B-y)\rvert $ exists and it is not zero at $y=0$. We conclude that $$\lvert B\cap (B-y)\rvert =\lvert B\rvert -C\lvert y \rvert + O(\lvert y\rvert^2), $$ which, by (1), gives $$\tag{2} \omega_{\chi_B}(y)= C\lvert y \rvert + O(\lvert y \rvert^2).$$ This is all we need, as it immediately implies that $$ \lVert \chi_B\rVert_{\dot{H}^s}^2= C \int_{\mathbb R^d} \frac{\omega_{\chi_B}(y)}{\lvert y\rvert^{d+2s}}\, dy <\infty \quad \iff \quad s<\frac12.$$

For an arbitrary $E\subset \mathbb R^d$ of finite measure, the result is that $$\lVert \chi_E\rVert_{\dot{H}^s}^2<\infty \quad \Longrightarrow \quad s<\frac12.$$ This follows from the above by the symmetric rearrangement, which gives $\lVert \chi_E\rVert_{\dot{H}^s}\ge C\lVert \chi_B\rVert_{\dot{H}^s}$, as cleverly shown by LL 3.14.

I have tried to bypass the symmetric rearrangement. The above argument would push through if we could show that $$ \omega_{\chi_E}(y)\ge C\lvert y \rvert + O(\lvert y \rvert^2), $$ but I couldn’t find a way to prove this. I don’t even know if this is true, actually.