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Consider a positive valued measurable function $f$ over a measure space $(\mathcal{X},\mathcal{M},\mu).$

Let $S(\mathcal{F})$ denote the linear span of characteristic functions of sets in the sigma algebra $\mathcal{F}.$

Is it true that

$$\sup_{0\leq\phi\leq f,\ \phi\in S(\mathcal{M})} \int_{\mathcal{X}}\phi d\mu = \sup_{0\leq\phi\leq f,\ \phi\in S(\sigma(f))} \int_{\mathcal{X}}\phi d\mu$$

where $\sigma(f),$ the minimal $\sigma-$algebra over which $f$ is measurable.

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    The k-cube used to integrate the manifold under some function but belong to the interior of the manifold for $ M \rightarrow \sigma(f_{k}) $ so $ I_{k} = \sum_{i =1}^{x_{n}} (c_{0,i} - c_{1,i}) \in M^{\circ} $ such that $ I_{k} \subseteq \chi $ Then the supremum of $S(M) = S(\sigma(f)) $ for $ \phi $ within the $\sigma$-algebra. – 2017-02-09
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    Yes. All you need is the fact that for each $\mathcal{X}$-measurable real-valued function $f$ you can find a pointwise increasing sequence $\langle f_n\rangle$ of $\mathcal{X}$-simple functions that converges pointwise to $f$, no matter which $\sigma$-algebra $\mathcal{X}$ is, and the monotone convergence theorem. Then it is clear that both sides are simply the Lebesgue integral of $f$. – 2017-02-09

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