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So the definition of Lebesgue integral as I understand it is as follows: Let $(X, \mathcal{F}, \mu)$ be a measure space, and $f: X \to [0, + \infty]$ a non-negative function. Then for simple functions, i.e. functions $\varphi(x) = \sum_{n = 1}^{N} c_{n} \chi_{E_{n}}(x)$ for some sets $E_{n} \in \mathcal{F}$, we let $\int_{E} \varphi(x) \mathrm{d} \mu (x) = \sum_{n = 1}^{N} c_{n} \mu( E_{n} \cap E)$, where $E \in \mathcal{F}$. We then define

\begin{align*} \int_{E} f(x) \mathrm{d} \mu (x) & := \sup \{ \int_{E} \varphi(x) \mathrm{d} \mu (x) : \varphi \textrm{ is simple}, \varphi(x) \leq f(x) \forall x \in E \} \end{align*}

We then define the integral for real-valued functions by considering their positive and negative parts, and then for complex-valued functions by considering the real and imaginary parts, and onward for $\mathbb{C}^{m}$-valued functions.

My question is if one could say equivalently

\begin{align*} \int_{E} f(x) \mathrm{d} \mu (x) & := \inf \{ \int_{E} \varphi(x) \mathrm{d} \mu (x) : \varphi \textrm{ is simple}, \varphi(x) \geq f(x) \forall x \in E \} \end{align*}

If not, then what would be a counter-example? Further, is there a sensible family of measure spaces for which the two are equivalent?

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No, that won't work. If $f$ is unbounded above, then every simple function $\varphi$ with $\varphi(x) \geq f(x)$ will be infinite on a set of positive measure, so your integral would be $+\infty$. Yet there are unbounded functions with finite integral, such as $f(x) = \frac{1}{\sqrt{x}} \chi_{[0,1]}(x)$.

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  • $\begingroup$ If I stuck to bounded functions, then would it work? $\endgroup$ – AJY Jan 28 '15 at 23:04
  • $\begingroup$ @AJY Yes, it'll work with bounded functions. (Also, would the downvoter care to comment?) $\endgroup$ – Ian Jan 28 '15 at 23:09

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