Lebesgue integral of f over R is bigger or equal to alpha times measure of $E_\alpha?$ Here is the problem: (1) Suppose that $f \geq 0$ is measurable and integrable. If $\alpha \geq 0$ and $E_α = \{x : f(x) > \alpha\},$ prove that $\lambda(E_\alpha) \leq \frac{1}{\alpha} \int{f(x) dx}.$
Now my question is, how can we have strict equality if my following proof is correct? I guess it isnt?
1) Let $f\geq 0$ be integrable, $\alpha>0, E_\alpha=\{x:f(x)>\alpha\} $
Since $f$ is measurable, $E_\alpha$ is measurable. $\int_{E_\alpha} f dx > \int_{E_\alpha} \alpha dx= \alpha m(E_\alpha) $
Hence $ \int_\mathbb{R} f = \int_{\mathbb{R}\setminus{E_\alpha}} f  + \int_{E_\alpha} f > \int_{\mathbb{R}\setminus{E_\alpha}} f  + \int_{E_\alpha} \alpha = \int_{\mathbb{R}\setminus{E_\alpha}} f  + \alpha m({E_\alpha})  $.
Since $f \geq 0$ then $\int_{R\setminus{E_\alpha}} {f} \geq 0 $
$=> \alpha m(E_\alpha)< \int_\mathbb{R} f $ and dividing by $\alpha$ gives the desired result.
Where is my mistake? Also maybe less importantly, where is the integrability of f used? Even if f is not integrable ie its integral is infinite, the inequality should hold?Thanks in advance guys.

Your '>' holds if the measure of $E_{\alpha}$ is positive. If not, you have $\geq$. If $f$ is not integrable the inequallity holds (every set has measure $\leq\infty$ by definition.
