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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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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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  • $\begingroup$ Then : $\int_x^{x + \delta} f(t) dt < \epsilon$ but this will prove continuity in F(x), How can conclude that $|f(x + \delta) - f(x)|<\epsilon$ ? $\endgroup$ – Bunny Oct 31 '12 at 0:38
  • $\begingroup$ Wait, are we showing that $F$ is continuous, or $f$? $\endgroup$ – Christopher A. Wong Oct 31 '12 at 0:40
  • $\begingroup$ to show integrals are well defined, I should show continuity in $f$ right? $\endgroup$ – Bunny Oct 31 '12 at 0:42
  • $\begingroup$ Ah, my apologies. My answer does not even remotely address your question then. Let me rewrite it. $\endgroup$ – Christopher A. Wong Oct 31 '12 at 0:45
  • $\begingroup$ @Bunny, unfortunately such a claim is impossible; for example, a function can have countably many jumps at $x_n$, such that $f(x_n^{+}) - f(x_n^{-}) = 1/n^2$; then if you sum all the jumps it remains bounded. $\endgroup$ – Christopher A. Wong Oct 31 '12 at 2:02
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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.

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