135 It follows that f (x0)−ε <f (x) for all x ∈B(x0;δ)∩D. Therefore, f is lower semicontinuous. The proof for the upper semicontinuous case is similar. □ For every a∈R, we also define La(f ) ={x ∈D: f (x) <a}=f − 1((−∞,a)) and Ua(f ) ={x ∈D: f (x) >a}=f − 1((a,∞)). Corollary 5.4.6 Let f : D→R. Then f is lower semicontinuous if and only if Ua(f ) is open inD for every a∈R. Similarly, f is upper semicontinuous if and only if La(f ) is open in Dfor every a∈R. Theorem 5.4.7 Let f : D→R. Then f is continuous if and only if for every a,b∈Rwith a<b, the set Oa,b ={x ∈D: a<f (x) <b}=f − 1((a,b)) is an open set inD. Proof: Suppose f is continuous. Then f is lower semicontinuous and upper semicontinuous. Fix a,b∈Rwitha<b. Then Oa,b =Lb ∩Ua. By Theorem 5.4.6, the set Oa,b is open since it is the intersection of two open sets La andUb. Let us prove the converse. We will only show that f is lower semicontinuous since the proof of upper semicontinuity is similar. For everya∈R, we have Ua(f ) ={x ∈D: f (x) >a}= [ n∈N f −1((a,a+n)) Thus, Ua(f ) is open in Das it is a union of open sets inD. Therefore, f is lower semicontinuous by Corollary 5.4.6. □ Exercises 5.4.1 Let f be the function given by f (x) =( x2, if x̸ =0; −1, if x =0. Prove that f is lower semicontinuous. 5.4.2 Let f be the function given by f (x) =( x2, if x̸ =0; 1, if x =0. Prove that f is upper semicontinuous.
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