[Math] Sum of two upper semi continuous functions is upper semi continuous

real-analysissemicontinuous-functions

Question is to prove that Sum of two upper semi continuous functions is upper semi continuous..

This is the very first time i am dealing with upper semi continuous functions..

the very first upper semi continuous function is characteristic functions of closed sets.

Let $\chi_{K_1}$ and $\chi_{K_2}$ be two upper semi contiuous functions..

Their sum is $(\chi_{K_1}+\chi_{K_2})(x):=\chi_{K_1}(x)+\chi_{K_2}(x)$

Consider $\{x: \chi_{K_1}+\chi_{K_2}(x) <r\}$ for arbitrary $r\in \mathbb{R}$

Suppose $r=0 $ then $\{x: \chi_{K_1}+\chi_{K_2}(x) <r\}$ is empty set and so is open

Suppose $r=2$ then $$\{x: \chi_{K_1}+\chi_{K_2}(x) <2\}=\{x: \chi_{K_1}+\chi_{K_2}(x) =0\}\cup\{x: \chi_{K_1}+\chi_{K_2}(x) =1\}$$

Where as $\{x: \chi_{K_1}+\chi_{K_2}(x) =0\}=\emptyset$

So $$\{x: \chi_{K_1}+\chi_{K_2}(x) <2\}=\{x: \chi_{K_1}+\chi_{K_2}(x) =1\}$$

But
$$\{x: \chi_{K_1}+\chi_{K_2}(x) =1\}=\{x: \chi_{K_1}(x)=0; \chi_{K_2}(x) =1\}\cup \{x: \chi_{K_1}(x)=1; \chi_{K_2}(x) =0\}$$
And
$$\{x: \chi_{K_1}(x)=0; \chi_{K_2}(x) =1\}=K^c_1\cap K_2$$

$$\{x: \chi_{K_1}(x)=1; \chi_{K_2}(x) =0\}=K_1\cap K^c_2$$

We see $$(K^c_1\cap K_2)\cup(K_1\cap K^c_2) $$

But all these are not making it any simple…

If this is not simple i can not expect for arbitrary semi continuous functions it will be simple…

Please give hints for this and not answer.. and then i will try for general semi continuous and edit this…

EDIT : My definition of upper semi continuous function is :

I say $f:X\rightarrow \mathbb{R}$ to be upper semi continuous if $\{x : f(x)<r\}$ is open for all $r\in \mathbb{R}$

Please suggest methods which uses this definition..

Best Answer

Suppose $X$ is a topological space, $f,g : X \longrightarrow \mathbb{R}$ are two upper semi-continuous functions.

Then, for all $x \in X$ and for all $\varepsilon >0$ there exist two open neighbourhoods of $x$, called $U,V$ such that $$ \forall y \in U, \ f(y) \leq f(x) + \varepsilon $$ $$ \forall y \in V, \ g(y) \leq g(x) + \varepsilon $$ hence $$ \forall y \in U \cap V, \ f(y) + g(y) \leq f(x)+g(x) +2 \varepsilon $$ this means that $f+g$ is upper semi-continuous.