[proofplan] A direct comparison argument: bound $f$ below by $\lambda \cdot \mathbb{1}_{\{f \geq \lambda\}}$, integrate both sides using monotonicity, and divide by $\lambda$. [/proofplan] [step:Bound $f$ below by $\lambda \cdot \mathbb{1}_{\{f \geq \lambda\}}$ and integrate] Since $f \geq \lambda \cdot \mathbb{1}_{\{f \geq \lambda\}}$ pointwise (on the set $\{f \geq \lambda\}$ the indicator equals $1$ and $f \geq \lambda$, while off that set the indicator is $0$), integrating both sides and using monotonicity of the integral gives \begin{align*} \int_E f \, d\mu \geq \int_E \lambda \cdot \mathbb{1}_{\{f \geq \lambda\}} \, d\mu = \lambda \cdot \mu(\{f \geq \lambda\}). \end{align*} Dividing by $\lambda > 0$ yields the result. [/step]