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Do the following identitities hold: Inf(S+T)=inf(S)+inf(T), where S and T are sets of nonnegative real numbers Also sup(S+T)=sup(S)+sup(T)

Bbbh
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  • I am trying to prove that the sum of two regular Borel measures is a regular Borel measure and if the above identities are true, so the regularity is proved. – Bbbh May 12 '13 at 17:14
  • What does $S + U$ mean? Set union, ${x + y \colon x \in S \wedge y \in U}$, something else? – vonbrand Mar 07 '14 at 14:39
  • @TuckerRapu: I'd say your linked question is not a duplicate; for sets the sup of the sum is equal to the sum of the sups, while for functions there's generally an inequality. – Andrew D. Hwang Mar 07 '14 at 15:02
  • @user86418 thanks. are you referring to http://math.stackexchange.com/q/703040/85079 –  Mar 09 '14 at 14:54
  • @TuckerRapu: No, that's not the post I meant. I may have meant the post for which this question is now marked as a duplicate (incorrectly, in my opinion), but there are so many variants scattered about the site that it's difficult to be certain. – Andrew D. Hwang Mar 09 '14 at 15:09

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It's true: Let $m = \inf \{S+T\}$. i.e.

$$m = \inf_{s \in S, t \in T}s + t $$

which implies that for $\epsilon > 0$, $\exists s \in S, t \in T$ s.t

$$m \leq s + t \leq m + \epsilon$$ $$\Rightarrow \inf_{s \in S} s +t \leq m + \epsilon$$ $$\Rightarrow \inf_{s \in S} s +\inf_{t \in T}t \leq m + \epsilon$$

As $\epsilon$ is arbitrary now, we have $$\Rightarrow \inf_{s \in S} s +\inf_{t \in T}t \leq m $$

Now for the reverse, let $m_1 = \inf_{s \in S} s$, $m_2 = \inf_{t \in T} t$

Then by similar arguments, $\forall \epsilon > 0$, $\exists s \in S, t \in T$ s.t $$ s \leq m_1 + \epsilon/2, t \leq m_2 + \epsilon/2$$

$$\Rightarrow s+t \leq m_1 + m_2 + \epsilon$$

$$\Rightarrow \inf_{s \in S, t \in T}s + t \leq m_1 + m_2 + \epsilon$$ Take $\epsilon$ to 0 to get the reverse. QED

Gautam Shenoy
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  • This doesn't address the possibility that $S$ or $T$ may be unbounded above (though that case is easily dealt with). – Cameron Buie May 12 '13 at 19:11
  • One way would be to take a bounded subset from each and claim that infimum of the whole set is in the bounded set. – Gautam Shenoy May 12 '13 at 19:15
  • Oh, there's no problem for the infimum, since $S$ and $T$ are certainly both bounded below by $0$. I mean that we need to address the possibility of unboundedness above when finding the supremum. We can't just swap $\inf$ for $\sup$ and turn inequalities around. – Cameron Buie May 12 '13 at 19:56
  • I agree, but as you said, a minor technicality. Let the OP decide if more clarification is needed. – Gautam Shenoy May 12 '13 at 20:02