A function $f$ is lower semicontinuous iff it is its own lower boundary.

Question

The following is an exercise from Bruckner's Real Analysis:

Prove that a function $f$ is lower semicontinuous if and only if it is its own lower boundary.

Definition of semicontinuous :

A function $f$ is called lower semicontinuous on $[a, b]$ if for every $α \in \mathbb{R}$ the set ${\{x : f(x) > α }\}$ is open.

Definition of lower boundary :

Let $f$ be a bounded function defined on $[a, b]$, let $x_0 \in [a, b]$, and let $δ > 0$. Write $m_δ(x_0)= \inf {\{f(x): x \in (x_0 − δ, x_0 + δ) \cap [a, b] }\}$ and define $m(x_0 ) = \lim_{δ→0} m_δ (x_0)$. The function $m$ is called the lower boundary of $f$.

My attempt : $(\Leftarrow)$ If $m(x_0) > α$, then there exists $β>α$ such that $f>β$ in a neighborhood $I$ of $x_0$, and hence $m>α$ on $I$. Thus ${\{x : m(x) > α}\}$ is open. So because $f=m$ then $f$ is lower semicontinuous.

My question : Am I right with one direction and how can I prove the other direction?

Answer 1

Let us prove that:

A function $f$ (defined in $[a,b]$) is lower semicontinuous if and only if it is its own lower boundary.

Proof:

$(\Rightarrow)$ Suppose $f$ is lower semicontinuous and let $x_0 \in [a,b]$. It is immediate that for all $\delta >0$, $$ m_\delta(x_0) = \inf {\{f(x): x \in (x_0 − \delta, x_0 + \delta ) \cap [a, b] }\} \leq f(x_0)$$

So

$$ m(x_0) = \lim_{\delta \to 0} m_\delta(x_0) \leq f(x_0) \tag{1} $$

Now, for any $\alpha \in \Bbb R$, such that $\alpha < f(x_0)$, we have that $x_0 \in {\{x : f(x) > \alpha }\}$ which is an open set. So there is $\delta >0$, such that $$(x_0 − \delta, x_0 + \delta) \cap [a, b] \subseteq {\{x : f(x) > \alpha }\}$$ So $m_\delta(x_0) \geq \alpha$.

It is easy to see that, for all $\delta_1, \delta_2 >0$, if $\delta_1 < \delta_2$ then $m_{\delta_1}(x_0) \geq m_{\delta_2}(x_0)$ (that means, $m_\delta(x_0)$ is non-increasing with respect to $\delta$). So $$ m(x_0) = \lim_{\delta \to 0} m_\delta(x_0) \geq \alpha$$

So, we have proved that, for all $\alpha \in \Bbb R$, such that $\alpha < f(x_0)$, we have $ m(x_0) \geq \alpha$. It follows that

$$ m(x_0) \geq f(x_0) \tag{2}$$

From $(1)$ and $(2)$, we have $ m(x_0) = f(x_0)$,

$(\Leftarrow)$ Suppose that, for all $x_0 \in [a,b]$, $ m(x_0) = f(x_0)$. Given any $\alpha \in \Bbb R$, let us prove that ${\{x : f(x) > \alpha }\}$ is open.

Consider any $x_0 \in {\{x : f(x) > \alpha }\}$. Then $ m(x_0) =f(x_0) > \alpha$.

Since, for all $\delta_1, \delta_2 >0$, if $\delta_1 < \delta_2$ then $m_{\delta_1}(x_0) \geq m_{\delta_2}(x_0)$, we have that, $m_\delta(x_0)\nearrow m(x_0)$ as $\delta \searrow 0$. So there is $\delta_0 >0$ such that for all $\delta < \delta_0, \alpha < m_\delta(x_0) \leq m(x_0)$.

But $\alpha < m_\delta(x_0)$ means that $$(x_0 − \delta, x_0 + \delta) \cap [a, b] \subseteq {\{x : f(x) > \alpha }\}$$ So, ${\{x : f(x) > \alpha }\}$ is open.