As the title states I wonder if it is true, and if so how to show, that $\lim_n n \int _c^{c+\frac {1 } {n }}f(x) \lambda(dx)=f(c)$ for bounded Lebesgue integrable $f $.
For a function continuous at $c $ this must be true since $\int _c^{c+\frac {1 } {n }}f(x) \lambda(dx)$ is about equal to $\frac {1 } {n } f(c) $.
For bounded Lebesgue integrable $f $ we have something like $1_ {[c,c +\frac {1 } {n }]}(x)f(x)\to 1_{\{c\} }(c) f(c)$, and we may use the Bounded convergence theorem, but I'm not sure how to treat this limit and the integral of it.
(The question originates from the proof on the following page http://mathonline.wikidot.com/the-derivative-of-functions-of-lebesgue-integrals)
Any help would be appreciated!
LHS does not change if you change the value of $f$ at $c$ to any number. So there is no hope of such a result for arbitrary $f$ and $c$. However, the result is true for almost all $c$. In fact it is true whenever $c$ is what is called a Lebesgue point of $f$. You can look at Lebesgue points and Lebesgue's Theorem in Rudin's RCA.