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Let $f$ be continous on $[a,b]$ and differentiable a.e. on $(a,b)$. Suppose there is a non-negative

function $g$ integrable on $[a,b]$ and $|\displaystyle \frac{f(x+1/n)-f(x)}{1/n}|\leq g$ a.e on $[a,b]$ for all $n$.

How can I show that: $\quad\displaystyle\int_a^bf'=f(b)-f(a). $

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@Bunny: Maybe you could use one of the convergence theorems (for integrals). Can you think of one that might help? –  Jesse Madnick Dec 3 '12 at 4:51

1 Answer 1

Using Dominated Convergence, and writing $F$ for an antiderivative of $f$, $$ \int_{[a,b]}f'=\int_{[a,b]}\lim_{n\to\infty}\frac{f(x+1/n)-f(x)}{1/n}=\lim_{n\to\infty}\int_{[a,b]}\frac{f(x+1/n)-f(x)}{1/n}\\ =\lim_{n\to\infty}n\int_{[a,b]}{f(x+1/n)-f(x)}=\lim_{n\to\infty}n\int_{a+1/n}^{b+1/n}{f(x)-n\int_a^bf(x)} \\ =\lim_{n\to\infty}n\int_{b}^{b+1/n}{f(x)-n\int_a^{a+1/n}f(x)} =\lim_{n\to\infty}\frac{F(b+1/n)-F(b)}{1/n}-\frac{F(a+1/n)-F(a)}{1/n}\\ =F'(b)-F'(a)=f(b)-f(a) $$

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Martin: How do you know that $f(x)$ is integrable on $[a+1/n, b+1/n]$? –  Anita Dec 12 '12 at 2:07
    
Note that by hypothesis, $f(x+1/n)-f(x)$ is integrable on $[a,b]$ (if you want, this is a defect of the question, as it does not mention that $f$ actually needs to be defined outside $[a,b]$). As $f$ is continuous on $[a,b]$ it is integrable there; so $f(x+1/n)$ is integrable on $[a,b]$. And $$\int_a^b|f(x+1/n)|dx=\int_{a+1/n}^{b+1/n}|f(x)|dx,$$ so $f$ is integrable on $[a+1/n,b+1/n]$. –  Martin Argerami Dec 12 '12 at 3:08

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