Suppose a continuos lebesgue measuable function $f$ that is (non negative) bounded above by $M$.
Define a sequence (similiar to standard represtation in counting measure) $\displaystyle f_m=\sum_{i=0}^{m-1}a_{i,m}\chi_{E_{i,m}}$ where $a_{i,m}=\frac{M}{m}i$ and $E_{i,m}=\{x|\frac{M}{m}i<f(x)<\frac{M}{m}(i+1) \}$.
It's obvious that as ${m\rightarrow \infty}, f_m$ in $L$ converges pointwise to $f$ and by Lebesgue Dominated Convergence Theorem $\int f d\mu=\lim_{m\rightarrow\infty} \int f_m d\mu$.
However, for a cointinuous function say $f(s)=e^{-|x|}$, the set $\lim_{m\rightarrow\infty}E_{i,m}$ become set of point elements. Thus $\mu(E_{i,m})=0$ for all i. Thus $\int fd\mu=0$ which is obvious not correct because riemann integration and Lebesgue integration are equivelent in this case.
What happend to the above process? How come $E_{i,m}$ has measure $0$ yet the lebague integration is non negative?
It's not quite right that $\mu(E_{i,m}) = 0$. What is right though, is that $\forall i, \lim_{m \rightarrow \infty} \mu(E_{i,m}) = 0$. So while you are indeed adding smaller and smaller numbers, you are adding more and more of them, as the sum defining $f_m$ goes from $0$ to $m$, and $m$ is growing to infinity. It's the very same principle as the Riemann sum converging to the Riemann integral.