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There is a classic Lemma about $sup A$ that says:

Let $A$ be any subset of the Real Numbers and let $s$ be an upper bound for $A$. $s = supA$ iff for every $\epsilon>0$ there exists $a \in A$ such that $s- \epsilon<a$. Similarly for $l=infA$ the inequality takes the form $a<l+\epsilon$.

In many books (e.g. Stein Real Analysis, p. 13), after define the outer measure of a set they use the above Lemma but instead of $a<l+\epsilon$ they write $a\leq l+\epsilon$.

Is this Correct??? and if yes How can I prove it.

Thanks!

Studentmathever
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1 Answers1

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Let $l$ be such that for every $\epsilon > 0$ there exists some $a\in A$ with $a \le l + \epsilon$.

Now, let $\tilde \epsilon > 0$. Then, for $\epsilon=\tilde\epsilon /2$ there exists some $a\in A$ with $$ a \le l + \tilde\epsilon /2 < l + \tilde\epsilon. $$

It is only important that $\epsilon$ is positive.

Note: You have forgotten to mention that $l$ should be a lower bound of $A$ :D

user251257
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