## Given least upper bound $\alpha$ for $\{\ f(x) : x \in [a,b] \ \}$, $\forall \epsilon > 0 \ \exists x$ s.t. $\alpha – f(x) < \epsilon$

I can’t figure out how all of this follows. Taken from Ch.8 of Spivak’s Calculus.

If $$\alpha$$ is the least upper bound of $$\{\ f(x) : x \in [a,b] \ \}$$ then, $$\forall \epsilon > 0 \ \exists x\in [a,b] \ \ \ \ \ \ \ \alpha – f(x) < \epsilon$$ This, in turn, means that $$\frac{1}{\epsilon} < \frac{1}{\alpha – f(x)}$$

## how to prove $\Phi \vdash \forall x.R(x,x)$

I’m completely stuck in part 3 of below exercise, and dont even know how to begin proving this, other than writing all the premises

## $f \in L^p$ for $1\le p c||f||_p\}\Big) \le \frac{1}{c^p} \ \forall c>0$

$$f \in L^p$$ for $$1\le p < \infty$$. The measure space is $$(\Omega, \mathcal{A}, \mu)$$.

How can I show that

$$\mu\Big(\{x \in \Omega:|f(x)|>c||f||_p\}\Big) \le \frac{1}{c^p} \ \forall c>0$$