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In reading the top-voted answer on this post, the answer appears to use the following fact (in the first bullet point of the answer):

Claim: If $f: [0,1] \to \mathbb{R}$ is of bounded variation, then $f'$ is absolutely integrable (i.e. $\int_{0}^{1} |f'(x)| dx < \infty$).

Is this true? If so, how can we prove it?

My proof attempt goes as follows: Suppose that $f$ is of bounded variation on $[0,1]$. Fix a partition $0 = t_0 < t_1 < \cdots < t_N = 1$ of $[0,1]$. Then $$ \sum_{j=1}^{N} |f(t_j) - f(t_{j-1})| \leq M$$ for some $M < \infty$ independent of the partition. Now I want to apply the Mean Value Theorem here, but I'm not sure if we can do this because $f'$ only exists almost everywhere. If $f'$ exists for all $x \in [0,1]$, then we can select some $c_j \in (t_{j-1},t_j)$ for each $j$ so that $ f(t_j) - f(t_{j-1}) = \sum_{j=1}^{n} |f'(c_j)|(t_j - t_{j-1})$, which gives us

$$ \sum_{j=1}^{N} |f'(c_j)| (t_j - t_{j-1}) \leq M, $$

which is the integral of the step function $\psi(x) = \sum_{j=1}^{N} |f'(c_j)| \chi_{(t_j - t_{j-1})}$. Now if we could choose a sequence of partitions so that the corresponding sequence of step functions increased to $f$, then I think the Monotone Convergence Theorem, together with the uniform bound $M$ for $\int \psi$, should finish the job. According to the accepted answer on this post, a function $f:I \to \mathbb{R}$ ($I$ a finite interval) is the limit of an increasing sequence of step functions if $f$ Riemann integrable, but I don't believe that being of bounded variation implies Riemann integrability (or does it?).

So to summarize my questions:

  1. Does the fact that $f'$ need not exist for all $x \in [0,1]$ ruin the above proof method? Or can the MVT still work?
  2. If $f$ is of bounded variation, can we say that $f$ is the limit of an increasing sequence of step functions?

Any other proofs of this claim would be appreciated!

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  • $\begingroup$ @MartinR: Thanks for pointing that out. Just fixed it. $\endgroup$
    – Leonidas
    Commented Feb 15, 2023 at 20:13
  • $\begingroup$ Is your question about the Riemann or Lebesgue integral? $\endgroup$
    – Martin R
    Commented Feb 15, 2023 at 20:14
  • $\begingroup$ The Lebesgue integral $\endgroup$
    – Leonidas
    Commented Feb 15, 2023 at 20:14
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    $\begingroup$ Check this: math.stackexchange.com/q/4609955/42969 $\endgroup$
    – Martin R
    Commented Feb 15, 2023 at 20:16
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    $\begingroup$ The fundamental fact here is that the derivative of a non decreasing function exists a.e. and is nonnegative so it has an integral whether finite or infinite and then the inequality $\int_a^bf'(x)dx \le f(b)-f(a)$ shows the integral is finite on any interval; a bounded variation function is a difference of such so $f=g-h, g,h$ nondecreasing implies $|f'|=g'+h'$ integrable $\endgroup$
    – Conrad
    Commented Feb 15, 2023 at 21:09

1 Answer 1

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Note that if $f:[0,1]\rightarrow\mathbb{R}$ is monotone increasing, then $f'$ exists a.e. and $f(1)-f(0)\geq\int_{0}^{1}f'\geq0$. In particular, $f'$ is integrable.

Now, if $g:[0,1]\rightarrow\mathbb{R}$ is of bounded variation, we may write $g=g_{1}-g_{2}$, for some increasing functions $g_{1}$,$g_{2}$ defined on $[0,1]$. Now that $g'=g_{1}'-g_{2}'$ a.e.. Since $g_{1}'$ and $g_{2}'$ are integrable, $|g'|$ is also integrable.

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