Riemann integrals and step functions

[GridironCPJ]

Prove the following:

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.

 

[GridironCPJ]

I've attached a PDF of my proof, which should be much easier to read. The converse is also proved in the following problem.View attachment Step Functions and Riemann Integrals.pdf

Your critiques would be helpful!

 

[math771]

Your proof looks correct to me. You have correctly used the definition of Riemann integrability and the fact that for any partition, the upper and lower sums can be made arbitrarily close to each other by choosing a small enough ε. You have also correctly defined the step functions L(x) and U(x) and shown that their difference is equal to the upper and lower sums. Your proof is clear and concise, and I do not see any flaws in your logic. Good job!