Sum of two upper semi continuous functions is upper semi continuous

Question

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..

Answer 1

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.