Suppose $f:[0,1] \to \mathbb{R}$ is continuous and has infinite total variation on any interval $[a,b] \subset [0,1]$ with $a <b$. Suppose $g$ is a function of bounded variation on $[0,1]$ and is continuous on $[0,1]$ i.e. $g \in BV([0,1]) \cap C([0,1])$. If $m$ denotes the Lebesgue measure on $\mathbb{R}$, is it true that $m(A)=0$ where $A \subset [0,1]$ is the set where $f$ and $g$ are equal?
I intuitively think that there are counterexamples using 'too bumpy' functions on fat cantor sets. But I am having a hard time constructing them rigorously. Can anyone help, please?
A way to construct such.
Pick $H$ your favorite continuous periodic function on $[0,1]$ that is not of BV on any subinterval. Any continuous function that is differentiable nowhere or on a set of measure zero will do since BV functions are differentiable ae on any interval.
Eg Riemann function $\sum_{n \ge 1} \frac{\sin 2\pi n^2x}{n^2}$ which is differentiable only at countably many points, or a Weierstrass function $G(2x)=\sum_{n \ge 0}(99/100)^n \cos (7^n 2\pi x)$ to make it of period $1$ and assume wlog (by adding a constant if needed) that $H(0)=H(1)=1$, while $|H(x)| \le M$ for $x \in \mathbb R$
For any $a<b$ let $H_{a,b}(x)=H(\frac {x-a}{b-a})$ the function with variables scaled so $H_{a,b}(a)=H_{a,b}(b)=1, |H_{a,b}| \le M, x \in [a,b]$ and of course $H_{a,b}$ is not of bounded variation on any subinterval of $[a,b]$
Consider a fat Cantor set, so a compact $C$ st $[0,1]-C=\cup (a_n, b_n), m(C)>0$ and $C$ has empty interior, and arrange the disjoint intervals $I_n=(a_n,b_n)$ in decreasing order of size, so at step $N$ we write $[0,1]=C_N \cup_{n \le N}I_N$, so if $(a_{jN}, b_{jN})$ are corresponding intervals $I_n$ in order from $0$ to $1$ now, we have $C_N=[0, a_{1N}]\cup [b_{1N}, a_{2N}]..\cup [b_{NN},1]$
Define two functions $f_N, g_N$ as follows; we first construct $F_N, G_N$ by taking a linear function from $0$ to $H_{a_{1N}, b_{a1N}}(a_{1N})=1$ on $[0, a_{1N}]$ then on $[a_{1N},b_{1N}]$ making the functions $F_N,G_N$ constant or $H_{a_{1N}, b_{1N}}(x)$ respectively, then making both increase linearly from $1$ to $2$ on the interval $[b_{1N}, a_{2N}]$ and then either take the constant $2$ or $H_{a_{2N}, b_{2N}}(x)+1$ on $[a_{2N},b_{2N}]$ etc while at the end take as $f_N=F_N/2^N, g_N=G_N/2^N$ and then take $f=\sum f_N, g=\sum g_N$
Note that by construction $|f_N| \le (N+1)/2^N, |g_N| \le (M+N)/2^N$ so clearly $f,g$ are well defined and continuous and $f(a_n)=g(a_n), f(b_n)=g(b_n)$ while $f$ is monotonic since each $f_N$ is so $f$ is BV and $f=g$ on the accumulation set of $a_n,b_n$ which is $C$ by definition, so $f=g$ on a set of positive measure.
But on each $(a_n,b_n)$ we have that for $N \ge n$ that $g_N(x)=\frac{H_{a_{n}, b_{n}}(x)+c_{nN}}{2^N}$, while for $N <n$ we have that $g_1,...g_{n-1}$ are increasing so $g$ restricted to $[a_n, b_n]$ is of the type $c_n+h_n(x)+d_nH_{a_{n}, b_{n}}(x)$ where $c_n, d_n >0$ are constants and $h_n$ is increasing, hence $g$ cannot have bounded variation on any subinterval of $[a_n,b_n]$
But now $C$ has empty interior so if $[a,b]$ is any subinterval of $[0,1]$ it cannot be included in it so it must have interior points in common with $(a_n,b_n)$ for some $n$ so it must have a subinterval in common with $(a_n,b_n)$ so it cannot be of bounded variation of $[a,b]$ since then it would be of bounded variation on a subinterval of $(a_n,b_n)$, so we are done!