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Let $f$ be an increasing function on $[0,1]$ . Let $F(x)= \int_0^x f(t) dt$. I want to show integrals are well defined. My attempts:

  1. $f$ is bounded, $f(0)\leq f(x)\leq f(1)$.

  2. $f$ may only have a countable set of jump since it's increasing.

How can I show that between jumps I have continuity?

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    See http://math.stackexchange.com/questions/156432/riemann-integral-problem2012-10-31

2 Answers 2

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Your attempt #2 is the right way to go about it. What you want to do is show that $f(x)$ only has a countable number of discontinuities (a.k.a. jumps). Then, by definition, any intervals between two discontinuities is continuous.

To do this, let $A_n = \{x \in [0,1]: f(x^{+}) - f(x^{-}) \geq 1/n\}$, where $f(x^{+}),f(x^{-})$ are the right and left side limits of $f$ at $x$, respectively. Can you show that this set is countable (or, indeed, finite)?

Then, to finish, the set $A = \bigcup_{n=1}^{\infty} A_n$ is precisely the set of all discontinuities.

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    How we can show that $A_n$ is a countable set?2012-10-31
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It is bounded since: $f(0)\leq f(x)\leq f(1)$

Denote for shorthand: $\mathcal{S}:=\{x:\Delta f(x)>0\}$

So countably many jumps: $\sum_{x\in\mathcal{S}}\Delta f(x)\leq f(1)-f(0)<\infty\implies\#\mathcal{S}\leq\#\mathbb{N}$

Thus it is integrable.