A continuous bounded variation function in $[0,1]$ that is absolutely continuous in $(a,1]$, but is not in $[0,1]$

I am seeking for a continuous of bounded variation function in $$[0,1]$$ that is absolutely continuous in $$(a,1]$$ for all $$a\in(0,1)$$, but is not in $$[0,1]$$.

The function $$x\sin\left(\frac{1}{x}\right)$$, for instance, is absolutely continuous and of bounded variation in $$(a,1]$$, but is not bounded in $$[0,1]$$.

• The latter function is continuous so it must be bounded, did you mean something else? Feb 19, 2022 at 20:27

There does not exist such a function. We recall that a function $$f:[a,b] \rightarrow \mathbb{R}$$ is absolutely continuous if and only if it is continuous, of bounded variation and has the property that $$f(A)$$ has Lebesgue measure $$0$$ whenever $$A\subset [0,1]$$ has Lebesgue measure $$0$$ (this last property is known as the Luzin-N-property). Now, if $$f:[0,1] \rightarrow \mathbb{R}$$ is as in your hypothesis and $$A\subset [0,1]$$ has Lebesgue measure $$0$$ then $$f(A)=f(A\cap \{0\})\cup \bigcup\limits_{n=1}^\infty f(A\cap [1/n, 1]).$$ The assumption that $$f$$ is a.c. when restricted to any subinterval $$(a, 1]$$, $$a>0$$, gives that all the sets in the union on the RHS have measure $$0$$, thus the LHS has measure $$0$$. Thus $$f$$ satisfies the Luzin-N-property and is a.c.
Edit: One can also prove it directly. Let $$\epsilon>0$$. We claim that there is a $$\delta>0$$ such that $$V_0^\delta (f)\leq\epsilon$$. Indeed, if this isn't the case we have $$V_0^\delta (f)>\epsilon$$ for all $$\delta>0$$. Using the continuity of $$f$$, pick $$\delta'>0$$ such that $$|f(x)-f(0)|<\epsilon/2$$ for $$x\in [0, \delta')$$. Then there is a partition $$0=x_{1, 0} with $$\sum\limits_{i=0}^{n_1-1}|f(x_{1,i+1})-f(x_{1,i})|>\epsilon$$. Then $$\sum\limits_{i=1}^{n_1-1}|f(x_{1,i+1})-f(x_{1,i})|>\epsilon/2.$$
Now use that $$V_0^{x_{1,1}} (f)>\epsilon$$ to obtain a partition $$0=x_{2, 0} with $$\sum\limits_{i=0}^{n_2-1}|f(x_{2,i+1})-f(x_{2,i})|>\epsilon$$. Then again $$\sum\limits_{i=1}^{n_2-1}|f(x_{2,i+1})-f(x_{2,i})|>\epsilon/2.$$ Continuing like this we obtain infinitely many non-overlapping partitions with total variation at least $$\epsilon/2$$. This contradicts that $$f$$ has bounded variation on $$[0,1]$$. The conclusion is that there is a $$\delta>0$$ such that $$V_0^\delta (f)\leq\epsilon$$. Since $$f$$ is a.c. on $$[\delta, 1]$$ there is a $$\delta''>0$$ witnessing the $$\epsilon$$ condition in absolutely continuity on $$[\delta, 1]$$ . Then $$\min(\delta, \delta'')$$ will witness the $$2\epsilon$$ condition in absolutely continuity on $$[0,1]$$.
• +1 Very nice. ${}$ Feb 19, 2022 at 20:39