# fundamental theorem of calculus for Riemann integration

- Let $f$ be a Riemann integrable function on an interval $[a,b]$ and $F$ defined in $[a,b]$ by $F(x)=\int_{a}^{x}f(t)\,dt+k$, where $k\in\mathbb{R}$ is a constant. Then $F$ is continuous in $[a,b]$ and $F^{\prime}=f$ almost everywhere (http://planetmath.org/MeasureZeroInMathbbRn).

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- Let $F$ be a continuous function in an interval $[a,b]$ and $f$ a Riemann integrable function such that $F^{\prime}(x)=f(x)$ except at most in a finite number of points $x$. Then $F(x)-F(a)=\int_{a}^{x}f(t)\,dt$.

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## 0.1 Observations

Notice that the second fundamental theorem is not a converse of the first. In the first we conclude that $F^{\prime}=f$ except in a set of measure zero (http://planetmath.org/MeasureZeroInMathbbRn), while in the second we assume that $F^{\prime}=f$ except in a finite number of points. In fact, the two theorems can never be the converse of each other as the following example shows:

Example : Let $F$ be the devil staircase function, defined on $[0,1]$. We have that

• $F$ is continuous in $[0,1]$,

• $F^{\prime}=0$ except in a set of (this set must be contained in the Cantor set),

• $f:=0$ is clearly a Riemann integrable function and $\int_{0}^{x}0\,dt=0$.

Thus, $F(x)\neq\int_{0}^{x}F^{\prime}(t)\,dt$.

This leads to the question: what kind functions $F$ can be expressed as $F(x)=F(a)+\int_{a}^{x}g(t)\,dt$, for some function $g$ ? The answer to this question lies in the concept of absolute continuity (http://planetmath.org/AbsolutelyContinuousFunction2) (a which the devil staircase does not possess), but for that a more general of integration must be developed (the Lebesgue integration (http://planetmath.org/Integral2)).

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