1. ## Supremum

Hi =)

I just want to check something...

If we have x and a given, and $\forall n \in \mathbb{N}$, $f_n(x) \leq a$ is this inequality true :
$\sup_n f_n(x) \leq a$
?

For those who are interested, it's part of a *simple* proof :
$(f_n)$ is a sequence of measurable functions : $(A, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$.
Prove that $\sup_n f_n$ is measurable.

$\square \quad \forall a \in \mathbb{R}, ~ \{\sup_n f_n \leq a\} \overset{(*)}{=} \bigcap_n \{f_n \leq a\}$

$\text{Note that } \{f_n \leq a \}=\{x \in \mathbb{R} ~:~ f_n(x) \leq a\}$

And $\{f_n \leq a \} \in \mathcal{A} \quad \square$

Thanks =)

2. That's correct. Weak inequalities are preserved by sup's (as are inf's and limits). But strict inequalities are not necessarily preserved by sup's (or by inf's or limits).

3. Originally Posted by Opalg
That's correct. Weak inequalities are preserved by sup's (as are inf's and limits). But strict inequalities are not necessarily preserved by sup's (or by inf's or limits).
Okay, thanks ^^

Is there a proof of that ? And an example for strict inequalities ?

4. Originally Posted by Moo
Okay, thanks ^^

Is there a proof of that ? And an example for strict inequalities ?
Proof by contradiction: suppose that $x_n\leqslant a$ for all n. If $s = \sup\{x_n\}>a$ then there must exist an x_n arbitrarily close to s. In particular, it can be made greater than a ...

Counterexample for strict inequalities: $1-\tfrac1n<1$. But $\sup(1-\tfrac1n) = 1$, which is not strictly less than 1.