# Essential supremum and strictly positive measure

Let $f: \mathbb{R}\to\mathbb{R}$ and the essential supremum of $f$ $$e=\inf\{\alpha\ge 0 \mid |f|\le\alpha\text{ almost everywhere}\}$$

I can't see why $$\lambda^*\left(\{x\in\mathbb{R} \mid |f(x)|\ge e-\epsilon\}\right)>0$$ for $a<b$ and $\epsilon>0$, where $\lambda$ is the Lebesgue measure.

Could someone help me ?

Fixed after Didier Piau's comment.

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The assertion is false. It is true if you replace $(a,b)$ by $\mathbb{R}$. –  Did Mar 27 '11 at 17:23
Or the statement should be: There are $a$ and $b$ such that $\ldots >0$. –  Rasmus Mar 27 '11 at 17:44
@Didier Piau: Thank you, I fixed the question. –  Klaus Mar 27 '11 at 18:17
if $\alpha<e$ then $|f|$ is not less or equal $\alpha$ a.e. (else $\alpha\geq e$ by definition), so there is a set of positive measure where $|f|>\alpha$.