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Let $f: [a,b] \to \mathbb R$. If $f$ is (Riemann-)integrable on $[a,b]$, then define $F: [a,b] \to \mathbb R$ by $F(x) = \int_a^x f.$

We have the following:

$\begin{array}{ccccccc} f \text{ differentiable} & \implies & f \text{ continuous} & \implies & f \text{ integrable} & \implies & ?_1 \\ \Big\Downarrow & & \Big\Downarrow & & \Big\Downarrow & & \Big\Downarrow \\ ?_2 & \implies & F \text{ differentiable} & \implies & F \text{ continuous} & \implies & F \text{ integrable} \end{array}$

Of course, we can define $\mathcal F(x) = \int_a^x F, \qquad \mathfrak F(x) = \int_a^x \mathcal F, \qquad \text{etc.}$ for this chain of implications to grow indefinitely.

My questions are

  1. ($?_1$) Is there a property of $f$, strictly weaker than integrability, that guarantees that $F$ is integrable?
  2. ($?_2$) Similar.

Edit: Ignore $?_1$. That was just plain stupidity on my part. If $f$ isn't integrable, then $F$ isn't even defined. Oops.

  • 0
    @LeonidKovalev, I am not familiar with the $C^k$ classes, so I have two questions: 1. Can you please elaborate on what you mean by "and conversely"? Are you saying $f$ continuous iff $F \in C^1$, etc.? 2. By "and so on" do you mean extending the chain diagram "leftwards" indefinitely?2012-06-19

2 Answers 2

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If $f$ is differentiable, then $F$ will be twice differentiable. I don't think you can get anything more than that.

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To get an $F$ that is not continuous, you would want to integrate not a function, but a measure. That is, if $\mu$ is a (signed) Borel measure, $F(x) = \int_{[a,x]} d\mu(t)$ is a right-continuous function of bounded variation (and in particular is Riemann integrable on bounded intervals).