Let $f:[a, b] \to \mathbb{R}$ and $\operatorname{Var}_{[a, b]} f$ its total variation. Let $\lambda$ be the Lebesgue measure on $[a, b]$. I would like to prove below properties for functions of bounded variation, i.e.,
Theorem If $f$ is of bounded variation, then
- $f$ is differentiable $\lambda$-a.e.,
- The function $$ g(x) := \begin{cases} f^{\prime}(x) & \text{if } f \text { differentiable at } x \in (a, b),\\ 0 & \text {otherwise}. \end{cases} $$ is $\lambda$-integrable such that $\int_a^b |g| \mathrm d \lambda \leq \operatorname{Var}_{[a, b]} f$.
Could you confirm if my below attempt is fine?
Proof
- By this lemma, there exist increasing functions $f_1, f_2:[a, b] \rightarrow \mathbb{R}$ such that $f=f_1-$ $f_2$ and $$ \operatorname{Var}_{[a, b]} f = f_1(b)-f_1(a)+f_2(b)-f_2(a). $$
By this lemma, $f_1, f_2$ are differentiable $\lambda$-a.e. Claim (1.) then follows.
- Let $$ g_1(x) := \begin{cases} f_1^{\prime}(x) & \text{if } f_1 \text { differentiable at } x \in (a, b),\\ 0 & \text {otherwise}. \end{cases} $$ and $$ g_2(x) := \begin{cases} f_2^{\prime}(x) & \text{if } f_2 \text { differentiable at } x \in (a, b),\\ 0 & \text {otherwise}. \end{cases} $$
I proved that $g, g_1, g_2$ are Borel measurable. So $g, g_1, g_2$ are $\lambda$-measurable. I proved that $g_1, g_2 \ge 0$ and $$ \int_a^b g_1 \mathrm d \lambda \le f_1(b)-f_1(a) \quad \text{and} \quad \int_a^b g_2 \mathrm d \lambda \le f_2(b)-f_2(a). $$
Notice that $g=g_1-g_2$ $\lambda$-a.e., so $$ \int_a^b |g| \mathrm d \lambda \le \int_a^b |g_1| \mathrm d \lambda + \int_a^b |g_2| \mathrm d \lambda = (f_1(b)-f_1(a))+(f_2(b)-f_2(a)) = \operatorname{Var}_{[a, b]} f < \infty. $$
This completes the proof.
