If $f$ is bounded Riemann integrable over $[a,b],$ then there is $\{P_n\}$ of partitions with $\lim_{n\to\infty}[U(f,P_n]-L(f,P_n)]=0.$

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The following question is taken from Real Analysis by Royden, Chapter $4,$ question $4:$

Suppose the bounded functoin $f$ on $[a,b]$ is Riemann integrable over $[a,b].$ Show that there is a sequence $\{P_n\}$ of partitions of $[a,b]$ for which $\lim_{n\to\infty}[U(f,P_n]-L(f,P_n)]=0.$

My attempt:

Fix a natural number $n.$ Then there exists two sequences of partitions $\{P_n'\}$ and $\{P_n''\}$ of $[a,b]$ such that $$\overline{\int}_a^b f-\frac{1}{n}\leq L(f,P_n') \leq \underline{\int}_a^b f = \overline{\int}_a^b f \leq U(f,P_n'') < \overline{\int}_a^b f + \frac{1}{n}.$$ Let $P_n = P_n'\cup P_n''.$ Then $P_n$ is a refinement of $P_n'$ and $P_n''.$ Therefore, $$L(f,P'_n) \leq L(f,P_n) \leq \overline{\int}_a^b f$$ and $$\overline{\int}_a^b f \leq U(f,P_n) \leq U(f,P_n'').$$ Hence, $$\lim_{n\to\infty}[U(f,P_n) - L(f,P_n) ] \leq \lim_{n\to\infty} \frac{2}{n} = 0.$$

Is my proof correct?