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A continuous extension of a vector field across one point need not be differentiable, so it does not automatically meet the classical hypotheses of the divergence theorem. Instead apply the divergence theorem after deleting a ball $B_{\epsilon }$ around that point. If $\Vert \mathbf{F}\Vert \le C{\epsilon }^{\alpha }$ on its boundary in $d$ dimensions, the magnitude of the inner flux is at most a constant times ${\epsilon }^{d-1+\alpha }$. When $d-1+\alpha >0$ this tends to zero. If the divergence is also integrable, passing to the limit recovers the outer flux from the volume integral. For example, a planar vector field with $\Vert \mathbf{F}\Vert \le r$ has an inner flux of order ${\epsilon }^2$, even when its extension is not differentiable at the origin.

For $r>0$, the partial derivatives $r_x=x/r$ and $r_y=y/r$ giveAdding these partial derivatives yields the divergence:The original vector field is undefined at the origin. Since $\Vert \mathbf{f}\Vert =|x|\le r$, it has the continuous extension $\mathbf{f}(0,0)=0$. This extension is not differentiable: on $y=0$ its first component is $|x|$, whose derivative at zero does not exist. Thus the displayed divergence has no classical value at the origin.

The triangle is $0\le y\le x\le 1$. In polar coordinates this becomesbecause the outer side is $x=r\cos \theta =1$. The vector field and the Jacobian determinant giveEvaluate the double integral component by component:Consequently,The origin causes no ambiguity in this double integral: the continuous extension from part (a) is bounded there, and changing a value at a single point does not change the integral.

We give the explicit point $\boxed{x=1/4}$. Repeatedly choose the left remaining third and then the right remaining third. After $2m$ choices, the resulting closed interval is $[L_m,R_m]$, whereThese formulas follow from a finite geometric series. They show $L_m<1/4<R_m$, and that both endpoints tend to $1/4$. The point $1/4$ is also strictly inside each intermediate left third: its distance from $L_m$ is one quarter of the current length. It therefore never enters any assigned closed middle third or its endpoints, so $1/4\in C$.

Endpoint values in the Cantor function construction obey the same left/right choices, with halving in place of thirding. ThusAll these endpoints are in the domain of the partial function: $L_m$ is the right endpoint of a middle third assigned at round $2m$, and $R_m$ is the left endpoint of a middle third assigned at round $2m-1$.

Given $\epsilon >0$, choose $m$ so that $(2/3)4^{-m}<\epsilon$, and put $\delta =9^{-m}/4$. If $t\notin C$ and $|t-1/4|<\delta$, then $L_m<t<R_m$. The assumed monotonicity givesConsequently $|f(t)-1/3|<\epsilon$. This proves directly, on the stated partial domain,There are defined points arbitrarily close to $1/4$, for example $L_m$ and $R_m$, so the limit of a function is not vacuous. No continuous extension has been assumed. The residual $C$ here omits the endpoints that the question assigns values to; it is the usual Cantor set with those assigned endpoints removed.