Measure of Lebesgue point

Steps:

1. Consider $\phi_r(x)$ to be $1$ on $[x-r,x+r]$ and $0$ outside of $[x-2r,x+2r]$ and connects linearly from $x-2r$ to $x-r$ and from $x+r$ to $x+2r$. (I know this is not exactly $C_c^\infty$ but it is easier to work with and one can always choose the "smoothed out" version.)
2. $\phi_r'(x)=\frac{1}{r}$ on $[x-2r,x-r]$ and $-\frac{1}{r}$ on $[x+r,x+2r]$. Heuristically, this converges to Dirac delta from above and below, i.e. $\delta_{x^-}-\delta_{x^+}$
3. $Du(\{x\})=\lim_{r\to0}\int_I\phi_r(x)d(Du)(x)=\lim_{r\to0}\int_I\phi'_r(x)u(x)dx=u(x^-)-u(x^+)$
4. By definition, $\forall \epsilon >0,\ \exists R>0$ s.t. $\forall r<R$ (expect for Lebesgue measure $0$ set), $|u(x-r)-u(x^-)|<\epsilon$. Similarly for the plus part with the same $R$!
5. Combine the definition of Lebesgue point, and $|u(x)-u(x^\pm)|\leq |u(x^\pm)-u(y)|+|u(y)-u(x)|$ to get

$$\frac{1}{2r}\int_{x-r}^{x+r}|u(y)-u(x)|dy\geq \frac{1}{2}(|u(x)-u(x^-)|-\epsilon)+\frac{1}{2}(|u(x)-u(x^+)|-\epsilon)\geq |u(x^-)-u(x^+)|-2\epsilon$$

Remark: I am not so fond of what I did in Step 2 and 3.

Your claim is true, here goes the proof following your main intuition. In the following WLOG $x=0$. Consider a compactly supported smooth function $\phi: \mathbb{R} \to [0,1]$ such that $\text{supp}(\phi)\subset B(0,2)$ and $\phi =1 $ on $B(0,1)$, such that $\|\phi'\|_{\infty} \leq 2$. We will use as test functions suitable scaled versions of $\phi$, namely we consider for all $k\in \mathbb{N},k\geq 1$ the function $\phi_k$ defined via $$ \phi_k (x) = \phi(kx) $$ $\forall x\in \mathbb{R}$. Clearly $\text{supp}(\phi_k) \subset B(0, \frac{2}{k})$ and $\phi_k = 1$ on $B(0,\frac{1}{k})$. Notice that for the functions $\phi_k$ it holds that $\phi_k'(x) = k\phi'(kx)$ and thus that $\|\phi_k'\|_\infty \leq 2k$. Now, following your idea, we compute: \begin{align*} \int_\mathbb{R} \phi_k \, dDu &= -\int_{\mathbb{R}} u(y)\phi_k'(y)\,dy \\&= -\int_\mathbb{R} (u(y)-u(0))\phi_k'(y)\,dy. \end{align*} where I have used that $\int_\mathbb{R} u(0) \phi_k'(y) \,dy = 0$ since $\phi_k$ has compact support. We now try to estimate the right-hand side of the previous equation. Since $\text{supp}(\phi_k) \subset B(0,\frac{1}{k})$, and recalling the bound $\|\phi'_k\|_{\infty} \leq 2k$ we have: \begin{align*} \left|\int_\mathbb{R} (u(y)-u(0))\phi'_k(y)\,dy\right| &\leq \int_{-\frac{1}{k}}^{\frac{1}{k}} |u(y)-u(0)||\phi'_k(y)| \, dy \\&\leq \int_{-\frac{1}{k}}^{\frac{1}{k}} |u(y)-u(0)|2k \, dy \\&= 4\frac{1}{\frac{2}{k}} \int_{-\frac{1}{k}}^{\frac{1}{k}} |u(y)-u(0)|\,dy \to_{k\to \infty} 0 \end{align*} where the last passage follows from the assumption that 0 is a Lebesgue point. Therefore we have obtained that $\left|\int_\mathbb{R} \phi_k \,dDu\right| \to 0$ At this stage, de dominated convergence theorem implies $$ 0 = \lim_{k\to \infty} \int_{\mathbb{R}} \phi_k \, dDu = \int_{\mathbb{R}}\chi_{\{0\}} \, dDu = Du(\{0\}). $$ You have just to recall the definition of total variation of a measure $\mu$, that is $$ |\mu(E)| = \sup\{\sum_{i=1}^\infty |\mu(E_i)|\mid \{E_i\}_1^\infty \text{ is a partition of } E\}. $$ Indeed, since the only partition of the singleton $\{0\}$ is $\{\{0\}\}$ you obtain $|Du|(\{0\}) = |Du(\{0\})| = 0$.

Suppose $\{x\}$ is an atom of $|Du|$. A simple way to arrive at a contradiction is to show that atoms of $Du$ (and hence $|Du|$) correspond to jump discontinuities of $u$, since jump discontinuities are never Lebesgue points. This should make sense, since the distributional derivative of the Heaviside function is a Dirac delta.

To see this, first note that we can assume $u$ is increasing, since $u = u_1 - u_2$ with $u_i$ increasing, and if $x$ is an atom for $Du$, it’s an atom for at least one of the $Du_i$.

Also, $\mu:=Du$ is the Lebesgue-Stieljes measure associated to $u$, since for any compactly supported smooth $v$, by integration by parts for Lebesgue-Stieljes measures,

$$\int v \ d\mu = -\int u \ dv = -\int uv’ \ dx$$

Therefore, using continuity of measure, the definition of Lebesgue-Stieljes measure, and the fact that BV functions have left and right limits, we see that

\begin{align*} \mu(\{x\}) &= \lim_{r\to 0}\mu((x-r,x+r]) \\ &= \lim_{r\to 0} (u(x+r) - u(x-r))\\ &= u(x^+) - u(x^-) > 0 \end{align*}

Therefore, $u$ has a jump at $x$.