# If $f:[0,1] \to \mathbb{R}$ is of bounded variation, is $|f'|$ is integrable?

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!

• @MartinR: Thanks for pointing that out. Just fixed it. Commented Feb 15, 2023 at 20:13
• Is your question about the Riemann or Lebesgue integral? Commented Feb 15, 2023 at 20:14
• The Lebesgue integral Commented Feb 15, 2023 at 20:14
• Check this: math.stackexchange.com/q/4609955/42969 Commented Feb 15, 2023 at 20:16
• 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 Commented Feb 15, 2023 at 21:09

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.