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I have started studying Lebesgue integration and I have a question regarding the Lebesgue integral.

In the Wikipedia entry on "Lebesgue integration" they define the Lebesgue integral as:

Let $f: \mathbb{R} \rightarrow \mathbb{R}^{+}$ be a positive real-valued function. $$ \int f d\mu = \int_{0}^{\infty}f^{*}(t)dt $$ where $f^{*}(t) = \mu(\{x |f(x) > t\})$.

The Lebesgue integration notes that I am studying define the Lebesgue integral of a positive measurable function as $$ \int f d \mu = \text{sup}\left\{ \int \phi d\mu :\; \phi \text{ is a simple function and } 0 \leq \phi \leq f \right\} $$ I want to know if this wiki definition is equivalent to the integral constructed from simple functions: if so, how can this be easily shown?

Daniele Tampieri
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2 Answers2

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A hint: break the Riemann integral into the limit of a summation, and it should be clear that they are the same. $f^*(t)dt$ is the area on the strip between two step functions that are a distence $dt$ apart.

Stella Biderman
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  • How would you show that that gives you $\int f d\mu = \int_{0}^{\infty}f^{*}(t)dt = \text{sup}{\int \phi d\mu: \phi \text{ is simple and } 0 \leq \phi \leq f }$ if we assume that $f$ is a positive measurable function? –  Mar 05 '14 at 18:56
  • Since $f^$ is continuous and can be taken to be non negative, measurable, there is a sequence of monotonically increasing measurable functions converging to $f^$. – Stella Biderman Mar 06 '14 at 07:34
  • Thanks for your response, I accept that you are saying that $f^{}$ is continuous and I know that for any positive measurable function we have a increasing sequence of simple functions which converges to it. But how exactly does that show that $\text{sup}{ \int \phi d\mu: \phi \text{ is simple and } 0 \leq \phi \leq f } = \int_{0}^{\infty}f^{}(t)dt$? –  Mar 20 '14 at 16:27
  • Given that there is a sequence that converges to $f^$, what's the only thing that sup could evaluate to? How far apart are $f$ and f^$? – Stella Biderman Mar 21 '14 at 12:44
  • What do you mean "how far are $f$ and $f^{*}$ apart?" I'm not getting your point there, could you expand on that? –  Mar 22 '14 at 17:06
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Define $\int f d\mu$ as usual. (I.e. Supremum of $\sum x \mu(s^{-1}(x))$ where $s$ is a simple measurable function such that $s≦f$.)

Note that $f^*$ is almost everywhere continuous since it is monotonically decreasing. Moreover, it is continuous. (Why?)

So $\int_0^\infty f^*(t)dt$ is well-defined where the integral here is Riemann.

Since $f^*$ is continuous, as Stella mentioned, it can be represented as a summation.

Moreover, do you know a theorem :"For every nonnegative measurable function, there exists a monotonically increasing sequence of simple measurable functions converging to that measurable function"?

Construction in that proof shows that the above summation is exactly the Lebegue integral of $f$.

John. p
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  • @ John.p Thanks for your response, I accept all those observations of $f^{}$ and I agree that it is Riemann integrable since it is a monotone function. But how exactly does that show that $\text{sup}{ \int \phi d\mu: \phi \text{ is simple and } 0 \leq \phi \leq f } = \int_{0}^{\infty}f^{}d\mu$? I am familiar with the proof of the final result but I don't see how that shows the equivalence. –  Mar 20 '14 at 16:57