Prove that a bounded function \( f:[a, b] \rightarrow \mathbb{R} \) is Riemann integrable if and only if for every \( \varepsilon>0 \) there exists a partition \( P \) of \( [a, b] \), which may depend on \( \varepsilon \), such that \( U(f ; P)-L(f ; P)<\varepsilon \).

To prove that a bounded function \( f: [a, b] \rightarrow \mathbb{R} \) is Riemann integrable if and only if for every \( \varepsilon > 0 \) there exists a partition \( P \) of \( [a, b] \) such that \( U(f; P) - L(f; P) < \varepsilon \), we will show both directions of the equivalence. ### (1) If \( f \) is Riemann integrable, then \( U(f; P) - L(f; P) < \varepsilon \) for some partition \( P \). Assume \( f \) is Riemann integrable on \( [a, b] \). By the definition of Riemann integrability, this means that the upper and lower sums can be made arbitrarily close to each other. Specifically, for any \( \varepsilon > 0 \), there exists a partition \( P \) of \( [a, b] \) such that: \[ U(f; P) - L(f; P) < \varepsilon. \] This is a direct consequence of the criterion for Riemann integrability, which states that a bounded function is Riemann integrable if and only if the set of discontinuities of \( f \) has measure zero. The closeness of the upper and lower sums can be achieved by refining the partition to account for the discontinuities. ### (2) If \( U(f; P) - L(f; P) < \varepsilon \) for some partition \( P \), then \( f \) is Riemann integrable. Now, assume that for every \( \varepsilon > 0 \), there exists a partition \( P \) such that: \[ U(f; P) - L(f; P) < \varepsilon. \] We need to show that \( f \) is Riemann integrable. To do this, we will show that the set of discontinuities of \( f \) has measure zero. Let \( D \) be the set of discontinuities of \( f \). For any \( \varepsilon > 0 \), we can find a partition \( P \) such that \( U(f; P) - L(f; P) < \varepsilon \). Now, consider the contribution to the upper and lower sums from the intervals in the partition \( P \). If we denote the intervals in the partition by \( [x_{i-1}, x_i] \), the upper sum \( U(f; P) \) is given by: \[ U(f; P) = \sum_{i=1}^{n} M_i (x_i - x_{i-1}), \] where \( M_i = \sup_{x \in [x_{i-1}, x_i]} f(x) \), and the lower sum \( L(f; P) \) is given by: \[ L(f; P) = \sum_{i=1}^{n} m_i (x_i - x_{i-1}), \] where \( m_i = \inf_{x \in [x_{i-1}, x_i]} f(x) \). The difference \( U(f; P) - L(f; P) \) can be interpreted as the total "error" introduced by the discontinuities of \( f \) over the intervals of the partition. If the set of discontinuities \( D \) had positive measure, we could find intervals in the partition that would contribute a significant amount to the difference \( U(f; P) - L(f; P) \), contradicting our assumption that this difference can be made less than \( \varepsilon \) for any \( \varepsilon > 0 \). Thus, the only way for \( U(f; P) - L(f; P) < \varepsilon \) to hold for every \( \varepsilon > 0 \) is if the set of discontinuities \( D \) has measure zero. Therefore, by the criterion for Riemann integrability, \( f \) is Riemann integrable. ### Conclusion We have shown both directions of the equivalence: 1. If \( f \) is Riemann integrable, then for every \( \varepsilon > 0 \), there exists a partition \( P \) such that \( U(f; P) - L(f; P) < \varepsilon \). 2. If for every \( \varepsilon > 0 \) there exists a partition \( P \) such that \( U(f; P) - L(f; P) < \varepsilon \), then \( f \) is Riemann integrable. Thus, we conclude that a bounded function \( f: [a,