Riemann integrals and step functions

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The discussion centers on proving that if a function f is Riemann integrable on an interval [a,b], then for any ε>0, there exist step functions L(x) and U(x) such that L(x) ≤ f(x) ≤ U(x) and the integral of their difference is less than ε. The proof references Theorem 8.16, indicating that a suitable partition π can be found to satisfy this condition. By defining L(x) as the infimum and U(x) as the supremum of f over each subinterval, the proof demonstrates that the sum of the differences between U(x) and L(x) over the partition is less than ε. The author seeks feedback on the proof's logic and clarity, noting that the details of Theorem 8.16 were included for context. Overall, the proof appears to effectively establish the required result.
GridironCPJ
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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.
 
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