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

In the wikipedia entry of "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 defines the Lebesgue integral of a positive measurable function as $$\int f d \mu = \text{sup}\{ \int \phi d\mu :\text{ } \phi \text{ is a simple function and } 0 \leq \phi \leq f \}$$ 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?


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 $f$.

  • $\begingroup$ @ 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. $\endgroup$ – user103184 Mar 20 '14 at 16:57

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.

  • $\begingroup$ 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? $\endgroup$ – user103184 Mar 5 '14 at 18:56
  • $\begingroup$ Since $f^*$ is continuous and can be taken to be non negative, measurable, there is a sequence of monotonically increasing measurable functions converging to $f^*$. $\endgroup$ – Stella Biderman Mar 6 '14 at 7:34
  • $\begingroup$ 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 $\text{sup}\{ \int \phi d\mu: \phi \text{ is simple and } 0 \leq \phi \leq f \} = \int_{0}^{\infty}f^{*}(t)dt$? $\endgroup$ – user103184 Mar 20 '14 at 16:27
  • $\begingroup$ 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^*$? $\endgroup$ – Stella Biderman Mar 21 '14 at 12:44
  • $\begingroup$ What do you mean "how far are $f$ and $f^{*}$ apart?" I'm not getting your point there, could you expand on that? $\endgroup$ – user103184 Mar 22 '14 at 17:06

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