91影视

To prove the given formula, let \(A_1,\ldots,A_m\) be an arbitrary partition of \(X\). Then, since \(f\) is bounded, let \(M=sup{f(x): x\in X}\). Now, we have: $$ \int f \mathrm{~d} \mu \leq \sum_{j=1}^{m} \mu\left(A_{j}\right) \sup_{A_j} f $$ Since the inequality holds for any arbitrary partition of \(X\), we can take the infimum over all possible partitions to get: $$ \int f \mathrm{~d} \mu \leq \inf \left\\{\sum_{j=1}^{m} \mu\left(A_{j}\right) \sup_{A_j} f: A_{1}, \ldots, A_{m} \text{ is an } \mathcal{S} \text{-partition of } X\right\\}. $$ Next, for any \(\epsilon>0\), we can find a partition \(B_1,\ldots,B_n\) of \(X\) such that: $$ \int f \mathrm{~d} \mu + \epsilon \geq \sum_{j=1}^{n} \mu\left(B_{j}\right) \sup_{B_j} f, $$ which can be achieved by choosing the partition such that \(\sup_{B_j} f \geq f(x) - \frac{\epsilon}{n\mu(X)}\) for all \(x\in B_j\). Taking the infimum again, we get: $$ \int f \mathrm{~d} \mu + \epsilon \geq \inf \left\\{\sum_{j=1}^{m} \mu\left(A_{j}\right) \sup_{A_j} f: A_{1}, \ldots, A_{m} \text{ is an } \mathcal{S} \text{-partition of } X\right\\}. $$ Since \(\epsilon>0\) is arbitrary, it implies: $$ \int f \mathrm{~d} \mu = \inf \left\\{\sum_{j=1}^{m} \mu\left(A_{j}\right) \sup_{A_j} f: A_{1}, \ldots, A_{m} \text{ is an } \mathcal{S} \text{-partition of } X\right\\}. $$ Therefore, part (a) is proven.

Let \((X, \mathcal{S}, \mu)\) be the measure space \(([0, 1], \mathcal{B}([0, 1]), \lambda)\), where \(\mathcal{B}([0,1])\) is the Borel sigma-algebra and \(\lambda\) is the Lebesgue measure. Consider the function \(f(x)=\frac{1}{\sqrt{x}}\) for \(x\in (0,1]\) and \(f(0)=0\). It is not bounded, but the integral is finite: $$ \int_0^1 f(x) \mathrm{~d} \lambda = 2 $$ Let's consider the expression: $$ \inf \left\\{\sum_{j=1}^{m} \lambda\left(A_{j}\right) \sup_{A_j} f: A_{1}, \ldots, A_{m} \text{ is a Borel } \text{-partition of } [0, 1]\right\\} $$ In any partition, there is some \(A_j\) containing a small interval around \(0\). Then, \(\sup_{A_j} f = \infty\) and thus: $$ \sum_{j=1}^{m} \lambda\left(A_{j}\right) \sup_{A_j} f = \infty $$ So, the infimum is also \(\infty\), and the expression doesn't match the finite integral of \(f(x)\), showing that the formula can fail when the boundedness condition is relaxed.

Consider the measure space \((\mathbb{R}, \mathcal{B}(\mathbb{R}), \lambda)\), where \(\mathcal{B}(\mathbb{R})\) is the Borel sigma-algebra and \(\lambda\) is the Lebesgue measure. Let \(f(x)=1\) for all \(x\in\mathbb{R}\). The function \(f(x)\) is bounded (specifically, \(M=1\)), but \(\mu(\mathbb{R}) = \infty\). Now, let's compute the integral of \(f(x)\) over \(\mathbb{R}\): $$ \int_{\mathbb{R}} f(x) \mathrm{~d} \lambda = \int_{-\infty}^{\infty} \mathrm{~d} x = \infty $$ In this case, the expression: $$ \inf \left\\{\sum_{j=1}^{m} \lambda\left(A_{j}\right) \sup_{A_j} f: A_{1}, \ldots, A_{m} \text{ is a Borel } \text{-partition of } \mathbb{R}\right\\} $$ is equal to the sum of the measures of all \(A_j\)'s, as the supremum of the constant function \(f(x)\) in each \(A_j\) is 1. However, since the measure space is not finite, the sum will also be infinite in this case, and the expression doesn't match the infinite integral of \(f(x)\), showing that the formula can fail when the condition of a finite measure space is removed.