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If f is Riemann integrable on an interval [a,b], show that ∀ε>0, there are a pair of step functions

L(x)≤f(x)≤U(x)

s.t.

∫_a^b▒(U(x)-L(x))dx<ε

My proof:

Since f is Riemann integrable on [a,b] then, by Theorem 8.16, ∀ε>0, there is at least one partition π of the interval [a,b] s.t.

∑_(k=1)^n▒〖ωf([x_(k-1),x_k ])(x_k-x_(k-1))〗<ε

Let L(x)=inf〖f(x)〗 ∀x∈[x_(k-1),x_k] and let U(x)=sup〖f(x)〗 ∀x∈[x_(k-1),x_k]. Note that

ωf([x_(k-1),x_k ])(x_k-x_(k-1) )=U(x)-L(x)

so

∑_(k=1)^n▒〖(U(x)-L(x))(x_k-x_(k-1))〗<ε

|∑_(k=1)^n▒〖(U(x)-L(x))(x_k-x_(k-1))〗-0|<ε

Hence,

∫_a^b▒(U(x)-L(x))dx<ε

Which is precisely the statement needed to be proven. ∎

I feel like my proof makes sense, but I would like to get some feedback to see if anyone sees any flaws in my logic. Note that I essentially explained the details of Theorem 8.16 in my proof. Also, ignore the ▒'s, as they were created when copying and pasting from Word.

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# Riemann integrals and step functions

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