Lebesgue integral of a positive function on a set of positive measure

Since $f$ is strictly positive on $E$, we have $$ E = \bigcup_{n \geq 1} E_n, \quad \mbox{ where } E_n = \left\{x \in E: f(x) > \frac{1}{n}\right\}. $$ Since $\lambda(E) > 0$ there is some $n$ for which $\lambda(E_n)$ is positive (otherwise $E$ would be the countable union of measure $0$ sets, implying $\lambda(E)=0$). We then have $$ \int_E f \, d\lambda \geq \int_{E_n} f \, d\lambda > \int_{E_n} \frac1n \, d\lambda = \frac{\lambda(E_n)}{n} > 0, $$ as desired.