A real-valued function \(f\) on a closed bounded interval \([a,b]\) is upper semicontinuous if for every \(\varepsilon>0\) and \(p\in[a,b]\) there exists \(\delta>0\) such that \[ x\in[a,b],\quad |x-p|<\delta \quad\Longrightarrow\quad f(x)<f(p)+\varepsilon. \] Prove that an upper-semicontinuous function is bounded above on \([a,b]\).
Solution
1. For every \(p\in[a,b]\), there is an open interval \(I_p\) containing\(p\) such that\[ x\in I_p\cap[a,b] \quad\Longrightarrow\quad f(x)<f(p)+1. \]
Apply upper semicontinuity at \(p\) with \[ \varepsilon=1. \] It gives \(\delta_p>0\) such that \[ \abs{x-p}<\delta_p \quad\Longrightarrow\quad f(x)<f(p)+1 \] for \(x\in[a,b]\). Take \[ I_p=(p-\delta_p,p+\delta_p). \]
2. There are points\[ p_1,\ldots,p_m\in[a,b] \]such that\[ [a,b]\subseteq I_{p_1}\cup\cdots\cup I_{p_m}. \]
The intervals \(I_p\), for \(p\in[a,b]\), form an open cover of the compact interval \([a,b]\). Compactness gives a finite subcover.
3. Define\[ M\coloneqq \max_{1\leq j\leq m}\bigl(f(p_j)+1\bigr). \]Then\[ f(x)<M \]for every \(x\in[a,b]\).
Fix \(x\in[a,b]\). By step 2, choose \(j\) such that \[ x\in I_{p_j}. \] Step 1 gives \[ f(x)<f(p_j)+1\leq M. \]
4. The function \(f\) is bounded above on \([a,b]\).
Step 3 gives the global upper bound \(M\).
5. Q.E.D.
Step 4 is the required conclusion.