Continuity

We are restricting ourself into $\mathbb{R}$ endowed with normal topology. Recall that a function is continuous if and only if for any open set $U \subset \mathbb{R}$, we have

$\{x:f(x) \in U\}=f^{-1}(U)$

to be open. One can rewrite this statement using $\varepsilon-\delta$ language. To say a function $f: \mathbb{R} \to \mathbb{R}$ continuous at $f(x)$, we mean for any $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$|f(x)-f(t)|<\varepsilon.$

$f$ is continuous on $\mathbb{R}$ if and only if $f$ is continuous at every point of $\mathbb{R}$.

If $(x-\delta,x+\delta)$ is replaced with $(x-\delta,x)$ or $(x,x+\delta)$, we get left continuous and right continuous, one of which plays an important role in probability theory.

But the problem is, sometimes continuity is too strong for being a restriction, but the ‘direction’ associated with left/right continuous functions are unnecessary as well. For example the function

$f(x)=\chi_{(0,1)}(x)$

is neither left nor right continuous (globally), but it is a thing. Left/right continuity is not a perfectly weakened version of continuity. We need something different.

Definition of semicontinuous

Let $f$ be a real (or extended-real) function on $\mathbb{R}$. The semicontinuity of $f$ is defined as follows.

If
$\{x:f(x)>\alpha\}$
is open for all real $\alpha$, we say $f$ is lower semicontinuous.
If
$\{x:f(x)<\alpha\}$
is open for all real $\alpha$, we say $f$ is upper semicontinuous.

Is it possible to rewrite these definition à la $\varepsilon-\delta$? The answer is yes if we restrict ourself in metric space.

$f: \mathbb{R} \to \mathbb{R}$ is upper semicontinuous at $x$ if for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have
$f(t)<f(x)+\varepsilon$
$f: \mathbb{R} \to \mathbb{R}$ is lower semicontinuous at $x$ if for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have
$f(t)>f(x)-\varepsilon$

Of course, $f$ is upper/lower semicontinuous on $\mathbb{R}$ if and only if it is so on every point of $\mathbb{R}$. One shall find no difference between the definitions in different styles.

Relation with continuous functions

Here is another way to see it. For the continuity of $f$, we are looking for arbitrary open subsets $V$ of $\mathbb{R}$, and $f^{-1}(V)$ is expected to be open. For the lower/upper semicontinuity of $f$, however, the open sets are restricted to be like $(\alpha,+\infty]$ and $[-\infty,\alpha)$. Since all open sets of $\mathbb{R}$ can be generated by the union or intersection of sets like $[-\infty,\alpha)$ and $(\beta,+\infty]$, we immediately get

$f$ is continuous if and only if $f$ is both upper semicontinuous and lower semicontinuous.

Proof. If $f$ is continuous, then for any $\alpha \in \mathbb{R}$, we see $[-\infty,\alpha)$ is open, and therefore

$f^{-1}([-\infty,\alpha))$

has to be open. The upper semicontinuity is proved. The lower semicontinuity of $f$ is proved in the same manner.

If $f$ is both upper and lower semicontinuous, we see

$f^{-1}((\alpha,\beta))=f^{-1}([-\infty,\beta)) \cap f^{-1}((\alpha,+\infty])$

is open. Since every open subset of $\mathbb{R}$ can be written as a countable union of segments of the above types, we see for any open subset $V$ of $\mathbb{R}$, $f^{-1}(V)$ is open. (If you have trouble with this part, it is recommended to review the definition of topology.) $\square$

Examples

There are two important examples.

If $E \subset \mathbb{R}$ is open, then $\chi_E$ is lower semicontinuous.
If $F \subset \mathbb{R}$ is closed, then $\chi_F$ is upper semicontinuous.

We will prove the first one. The second one follows in the same manner of course. For $\alpha<0$, the set $A=\chi_E^{-1}((\alpha,+\infty])$ is equal to $\mathbb{R}$, which is open. For $\alpha \geq 1$, since $\chi_E \leq 1$, we see $A=\varnothing$. For $0 \leq \alpha < 1$ however, the set of $x$ where $\chi_E>\alpha$ has to be $E$, which is still open.

When checking the semicontinuity of a function, we check from bottom to top or top to bottom. The function $\chi_E$ is defined by

$\chi_E(x)=\begin{cases}1 \quad x \in E \\0 \quad x \notin E\end{cases}.$

Addition of semicontinuous functions

If $f_1$ and $f_2$ are upper/lower semicontinuous, then so is $f_1+f_2$.

Proof. We are going to prove this using different tools. Suppose now both $f_1$ and $f_2$ are upper semicontinuous. For $\varepsilon>0$, there exists some $\delta_1>0$ and $\delta_2>0$ such that

$f_1(t) < f_1(x)+\varepsilon/2 \quad t \in (x-\delta_1,x+\delta_1), \\f_2(t) < f_2(x) + \varepsilon/2 \quad t \in (x-\delta_2,x+\delta_2).$

Proof. If we pick $\delta=\min(\delta_1,\delta_2)$, then we see for all $t \in (x-\delta,x+\delta)$, we have

$f_1(t)+f_2(t)<f_1(x)+f_2(x)+\varepsilon.$

The upper semicontinuity of $f_1+f_2$ is proved by considering all $x \in \mathbb{R}$.

Now suppose both $f_1$ and $f_2$ are lower semicontinuous. We have a identity by

$\{x:f_1+f_2>\alpha\}=\bigcup_{\beta\in\mathbb{R}}\{x:f_1>\beta\}\cap\{x:f_2>\alpha-\beta\}.$

The set on the right side is always open. Hence $f_1+f_2$ is lower semicontinuous. $\square$

However, when there are infinite many semicontinuous functions, things are different.

Let $\{f_n\}$ be a sequence of nonnegative functions on $\mathbb{R}$, then
If each $f_n$ is lower semicontinuous, then so is $\sum_{1}^{\infty}f_n$.
If each $f_n$ is upper semicontinuous, then $\sum_{1}^{\infty}f_n$ is not necessarily upper semicontinuous.

Proof. To prove this we are still using the properties of open sets. Put $g_n=\sum_{1}^{n}f_k$. Now suppose all $f_k$ are lower. Since $g_n$ is a finite sum of lower functions, we see each $g_n$ is lower. Let $f=\sum_{n}f_n$. As $f_k$ are non-negative, we see $f(x)>\alpha$ if and only if there exists some $n_0$ such that $g_{n_0}(x)>\alpha$. Therefore

$\{x:f(x)>\alpha\}=\bigcup_{n \geq n_0}\{x:g_n>\alpha\}.$

The set on the right hand is open already.

For the upper semicontinuity, it suffices to give an counterexample, but before that, we shall give the motivation.

As said, the characteristic function of a closed set is upper semicontinuous. Suppose $\{E_n\}$ is a sequence of almost disjoint closed set, then $E=\cup_{n\geq 1}E_n$ is not necessarily closed, therefore $\chi_E=\sum\chi_{E_n}$ (a.e.) is not necessarily upper semicontinuous. Now we give a concrete example. Put $f_0=\chi_{[1,+\infty]}$ and $f_n=\chi_{E_n}$ for $n \geq 1$ where

$E_n=\{x:\frac{1}{1+n} \leq x \leq \frac{1}{n}\}.$

For $x > 0$, we have $f=\sum_nf_n \geq 1$. Meanwhile, $f^{-1}([-\infty,1))=[-\infty,0]$, which is not open. $\square$

Notice that $f$ can be defined on any topological space here.

Maximum and minimum

There is one fact we already know about continuous functions.

If $X$ is compact, $f: X \to \mathbb{R}$ is continuous, then there exists some $a,b \in X$ such that $f(a)=\min f(X)$, $f(b)=\max f(X)$.

In fact, $f(X)$ is compact still. But for semicontinuous functions, things will be different but reasonable. For upper semicontinuous functions, we have the following fact.

If $X$ is compact and $f: X \to (-\infty,+\infty)$ is upper semicontinuous, then there exists some $a \in X$ such that $f(a)=\max f(X)$.

Notice that $X$ is not assumed to hold any other topological property. It can be Hausdorff or Lindelöf, but we are not asking for restrictions like this. The only property we will be using is that every open cover of $X$ has a finite subcover. Of course, one can replace $X$ with any compact subset of $\mathbb{R}$, for example, $[a,b]$.

Proof. Put $\alpha=\sup f(X)$, and define

$E_n=\{x:f(x)<\alpha-\frac{1}{n}\}.$

If $f$ attains no maximum, then for any $x \in X$, there exists some $n \geq 1$ such that $f(x)<\alpha-\frac{1}{n}$. That is, $x \in E_n$ for some $n$. Therefore $\bigcup_{n \geq 1}E_n$ covers $X$. But this cover has no finite subcover of $X$. A contradiction since $X$ is compact. $\square$

Approximating integrable functions

This is a comprehensive application of several properties of semicontinuity.

(Vitali–Carathéodory theorem) Suppose $f \in L^1(\mathbb{R})$, where $f$ is real-valued function. For $\varepsilon>0$, there exists some functions $u$ and $v$ on $\mathbb{R}$ such that $u \leq f \leq v$, $u$ is a upper semicontinuous functions bounded above, and $v$ is lower semicontinuous bounded below, and
$\boxed{\int_{\mathbb{R}}(v-u)dm<\varepsilon}$

It suffice to prove this theorem for $f \geq 0$ (of course $f$ is not identically equal to $0$ since this case is trivial). Since $f$ is the pointwise limit of an increasing sequence of simple functions $s_n$, we are able to write $f$ as

$f=s_1+\sum_{n=2}^{\infty}(s_n-s_{n-1}).$

By putting $t_1=s_1$, $t_n=s_n-s_{n-1}$ for $n \geq 2$, we get $f=\sum_n t_n$. We are able to write $f$ as

$f=\sum_{k=1}^{\infty}c_k\chi_{E_k}$

where $E_k$ is measurable for all $k$. Also we have

$\int_X f d\mu = \sum_{k=1}^{\infty}c_km(E_k),$

and the series on the right hand converges(since $f \in L^1$. By the properties of Lebesgue measure, there exists a compact set $F_k$ and a open set $V_k$ such that $F_k \subset E_k \subset V_k$ and $c_km(V_k-F_k)<\frac{\varepsilon}{2^{k+1}}$. Put

$v=\sum_{k=1}^{\infty}c_k\chi_{V_k},\quad u=\sum_{k=1}^{N}c_k\chi_{F_k}$

(now you can see $v$ is lower semicontinuous and $u$ is upper semicontinuous). The $N$ is chosen in such a way that

$\sum_{k=N+1}^{\infty}c_km(E_K)<\frac{\varepsilon}{2}.$

Since $V_k \supset E_k$, we have $\chi_{V_k} \geq \chi_{E_k}$. Therefore $v \geq f$. Similarly, $f \geq u$. Now we need to check the desired integral inequality. A simple recombination shows that

$\begin{aligned}v-u&=\sum_{k=1}^{\infty}c_k\chi_{V_k}-\sum_{k=1}^{N}c_k\chi_{F_k} \\ &\leq \sum_{k=1}^{\infty}c_k\chi_{V_k}-\sum_{k=1}^{N}c_k\chi_{F_k}+\sum_{k=N+1}^{\infty}c_k(\chi_{E_k}-\chi_{F_k}) \\ &=\sum_{k=1}^{\infty}c_k(\chi_{V_k}-\chi_{F_k})+\sum_{k=N+1}^{\infty}c_k\chi_{E_k}.\end{aligned}.$

If we integrate the function above, we get

$\begin{aligned}\int_{\mathbb{R}}(v-u)dm &\leq \sum_{k=1}^{\infty}c_k\mu(V_k-E_k)+\sum_{k=N+1}^{\infty}c_k\chi_{E_k} \\&< \sum_{k=1}^{\infty}\frac{\varepsilon}{2^{k+1}}+\frac{\varepsilon}{2} \\&=\varepsilon.\end{aligned}$

This proved the case when $f \geq 0$. In the general case, we write $f=f^{+}-f^{-}$. Attach the semicontinuous functions to $f^{+}$ and $f^{-}$ respectively by $u_1 \leq f^{+} \leq v_1$ and $u_2 \leq f^{-} \leq v_2$. Put $u=u_1-v_2$, $v=v_1-u_2$. As we can see, $u$ is upper semicontinuous and $v$ is lower semicontinuous. Also, $u \leq f \leq v$ with the desired property since

$\int_\mathbb{R}(v-u)dm=\int_\mathbb{R}(v_1-u_1)dm+\int_\mathbb{R}(v_2-u_2)dm<2\varepsilon,$

and the theorem follows. $\square$

Generalization

Indeed, the only unique property about measure used is the existence of $F_k$ and $V_k$. The domain $\mathbb{R}$ here can be replaced with $\mathbb{R}^k$ for $1 \leq k < \infty$, and $m$ be replaced with the respective $m_k$. Much more generally, the domain can be replaced by any locally compact Hausdorff space $X$, and the measure by any measure associated with Riesz-Markov-Kakutani representation theorem on $C_c(X)$.

Is the reverse approximation always possible?

The answer is no. Consider the fat Cantor set $K$, which has Lebesgue measure $\frac{1}{2}$. We shall show that $\chi_K$ can not be approximated below by a lower semicontinuous function.

If $v$ is a lower semicontinuous function such that $v \leq \chi_K$, then $v \leq 0$.

Proof. Consider the set $V=v^{-1}((0,1])=v^{-1}((0,+\infty))$. Since $v \leq \chi_K$, we have $V \subset K$. We will show that $V$ has to be empty.

Pick $t \in V$. Since $V$ is open, there exists some neighborhood $U$ containing $t$ such that $U \subset V$. But $U=\varnothing$ since $U \subset K$ and $K$ has empty interior. Therefore $V = \varnothing$. That is, $v \leq 0$ for all $x$. $\square$

Suppose $u$ is any upper semicontinuous function such that $u \geq f$. For $\varepsilon=\frac{1}{2}$, we have

$\int_{\mathbb{R}}(u-v)dm \geq \int_\mathbb{R}(f-v)dm \geq \frac{1}{2}.$

This example shows that there exists some integrable functions that are not able to reversely approximated in the sense of the Vitali–Carathéodory theorem.