# Taking the limit under the integral sign. Help!

Let f be continuous on [a,b] and differentiable almost everywhere on (a,b). Show that $$\int_a^b f'(x)\,\mathrm{d}x=f(b)-f(a)$$ if and only if $$\int_a^b\lim_{n\to\infty}\text{Diff}_{1/n}f=\lim_{n\to\infty}\int_a^b\text{Diff}_{1/n}f$$

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What is $\mathrm{Diff}$? –  martini Nov 23 '12 at 8:04
martini:Assume f is continuous. Extend f to take the value f(b) on (b,b+1] and for 0<h<=1, define the divided difference function on [a,b] by Diff_h f: $Diff_h f(x)=[f(x+h)-f(x)]/h$ for all x in [a,b]. –  i_a_n Nov 23 '12 at 8:14
$\def\D{\operatorname{Diff}}$If I should guess, I'd say $\D_h$ is the diffrence quotient operator. We have \begin{align*} \int_a^b \D_{1/n}f(x) \,dx &= \int_a^b \frac{f(x+1/n) - f(x)}{1/n} \, dx\\ &= n \cdot \left(\int_a^b f(x+1/n)\, dx - \int_a^b f(x)\, dx \right)\\ &= n \cdot \left(\int_{a+1/n}^{b+1/n} f(x)\, dx - \int_a^b f(x)\, dx\right)\\ &= n \cdot\int_b^{b+1/n} f(x)\, dx - n \cdot \int_a^{a+1/n} f(x)\, dx \end{align*} Now by continuity of $f$, for each $n$ there is an $a \le x_n \le a+1/n$ with $\int_a^{a+1/n} f(x)\, dx = \frac 1n f(x_n)$, analogously for the $b$-integral, again by continuity $n\int_a^{a+1/n} f(x)\, dx \to f(a)$, hence we have $$\int_a^b \D_{1/n}f(x) \, dx \to f(b) - f(a)$$ The limit on the left hand side is almost every where equal to $f'$ by definition. So your second equation is equivalent to $$\int_a^b f'(x)\, dx = f(b) - f(a)$$ which is what we wanted to show.
why by continuity of f, for each n there is an $a≤x_n≤a+1/n$ with $∫(a,a+1/n)f(x)dx=(1/n)*f(x_n)$ and $n∫(a,a+1/n)f(x)dx→f(a)$. Are these by any theorem? –  i_a_n Nov 24 '12 at 6:32
As $f$ is continuous, there are $y_i, y_s \in [a, a+1/n]$ with $f(y_i) \le f(x) \le f(y_s)$ for all $x \in [a,a+1/n]$. Integrating gives $f(y_i) \le n\int_a^{a+1/n}f(x)\, dx \le f(y_s)$. Now $x_n$ exists by the intermediate value theorem. Note that $x_n \to a$ for $n \to \infty$, hence $f(x_n) \to f(a)$ by continuity. –  martini Nov 24 '12 at 9:49