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Wikipedia presents a definition of the Lebesgue integral (of a nonnegative function $f$) that I hadn't encountered before:

Let $f^*(t)=\mu \left (\{x\mid f(x)>t\} \right )$.
The Lebesgue integral of $f$ is then defined by

$\int f\,d\mu = \int_0^\infty f^*(t)\,dt$
where the integral on the right is an ordinary improper Riemann integral

My question is, what is the relation between this definition and more standard definitions, like the supremum of $\int \phi\,d\mu$ over simple functions $\phi$ such that $0 \leq \phi \leq f$?

I'm also interested in understanding the intuitive justification provided for this definition:

Using the "partitioning the range of $f$" philosophy, the integral of $f$ should be the sum over $t$ of the area of the thin horizontal strip between $y = t$ and $y = t + dt$. This area is just

$\mu \left (\{x\mid f(x)>t\} \right ) \,dt$.

I don't see why that's the area of the infinitesimal strip. Clearly the width of the strip is $dt$, but why is the length of the strip $\mu \left (\{x\mid f(x)>t\} \right ) \,$?

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: It seems to me that the same reasoning that shows that the Lebesgue integral of $f$ can be expressed as $\int_0^\infty f^*(t)\,dt$ can also be used to express $f$ itself as $f(x) = \int_0^\infty f^{**}(x,t)\,dt$, where $f^{**}(x,t) = \chi_{\{s:f(s)>t\}}(x)$. Am I right about that?

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  • $\begingroup$ Tell me if I'm wrong, but I think this particular characterization of the Lebesgue integral is mentioned far more often when the subject is probability than when it's anything else. $\endgroup$ – Michael Hardy Apr 23 '14 at 18:01
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Say you have a nonnegative function $f$. Let $M > 0$ be fixed and let $n \in \mathbb N$. Partition the range of $f$ into the sets

  • $\left\{\dfrac{kM}{2^n} < f \le \dfrac{(k+1)M}{2^n}\right\}$ for $0 \le k \le 2^n - 1$, and
  • $\{f > M\}$.

Approximate the integral of $f$ by $$ \sum_{k=0}^{2^n - 1} \frac{kM}{2^n} \mu \left( \left\{ \frac{kM}{2^n} < f \le \frac{(k+1)M}{2^n} \right\} \right) + M \mu(\{f > M\}).$$

It isn't hard to see that this expression approximates the integral of $f$ because if $f$ is integrable then $M \mu(\{f > M\}) \to 0$ as $M \to \infty$, and the simple function $$\sum_{k=0}^{ 2^n - 1} \frac{kM}{2^n} \chi_{\{\frac{kM}{2^n} < f \le \frac{(k+1)M}{2^n}\}}(x)$$ approximates $f$ uniformly on the set $\{f \le M\}$.

Define $a_k = \dfrac{kM}{2^n}$ and $b_k = \displaystyle \mu \left( \left\{ f > \frac{kM}{2^n} \right\} \right)$. The sum above may be written as $$\sum_{k=0}^{2^n - 1} a_k (b_k - b_{k+1}) + a_{2^n} b_{2^n}.$$ Now employ the summation-by-parts trick to find this equal to $$ \sum_{k=1}^{2^n} (a_k - a_{k-1}) b_k = \sum_{k=1}^{2^n} \frac{M}{2^n} \mu( \{f > kM/2^n\}).$$As $n \to \infty$, the latter integral converges to $$\int_0^M \mu(\{f > t\}) \, dt.$$ Finally let $M \to \infty$ to get the integral of $f$.

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  • $\begingroup$ Thanks. Do you have any thoughts on Wikipedia's intuitive justification of the definition? Why would the area of the infinitesimal horizontal strip between $t$ and $t+dt$ be $\mu \left (\{x\mid f(x)>t\} \right ) \,dt$? $\endgroup$ – Keshav Srinivasan Apr 23 '14 at 18:31
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Umberto P.'s proof is quite rigorous, but I thought I'd try to give a slightly more intuitive (but less rigorous) explanation of how the Wikipedia definition follows from more common ways of thinking about the Lebesgue integral.

Let's partition the range $[0,\infty)$ into $y_1,y_2,y_3,...$ (where $y_1=0$), and let's chooser a point $y_i^*$ from each subinterval $[y_i,y_{i+1}]$. Then we can approximate our function $f$ by assuming that all points $x$ which map to the interval $[y_i,y_{i+1}]$ only map to the point $y_i^*$, so that we're approximating $f$ by the simple function $\sum_{i=1}^{\infty} y_i^*\chi_{\{x:y_i\leq f(x)\leq y_{i+1}\}}$, and thus approximating the Lebesgue integral of $f$ by $\sum_{i=1}^{\infty} y_i^*\mu({\{x:y_i\leq f(x)\leq y_{i+1}\}})$.

Now notice that ${\{x:y_i\leq f(x)\leq y_{i+1}\}}$ is the set of points for which $f(x)$ is between $y_i$ and $y_{i+1}$, or to put it another way $f(x)$ is greater than $y_i$ but is not greater than $y_{i+1}$, so we can express the set as ${\{x: f(x) > y_i\}}-{\{x: f(x) > y_{i+1}\}}$. Thus $\mu({\{x:y_i\leq f(x)\leq y_{i+1}\}})=\mu({\{x: f(x) > y_i\}})-\mu({\{x: f(x) > y_{i+1}\}})$, which by definition is equal to $f^*(y_i)-f^*(y_{i+1})$. (Note: I'm playing fast and loose with the distinction between less than and less than or equal to.)

So we can rewrite our approximation of the Lebesgue integral as $\sum_{i=1}^{\infty} y_i^*(f^*(y_i)-f^*(y_{i+1})) = -\sum_{i=1}^{\infty} y_i^*(f^*(y_{i+1})-f^*(y_i))$. Now we can see that in the limit as the width of the intervals $[y_i,y_i+1]$ goes to zero (i.e. the mesh of the partition goes to zero), this expression becomes the Riemann-Stieltjes integral $-\int_0^\infty ydf^{*}(y)$. Since $f^*$ is monotone, it's of bounded variation, so we're allowed to apply integration by parts to it, so we get $-(\lim_{y\rightarrow\infty}(yf^*(y))-0f^*(0)-\int_0^\infty f^*(y)dy)$. I think we can show that if the Lebesgue integral of $f$ exists, then $\lim_{y\rightarrow\infty}(yf^*(y))=\lim_{y\rightarrow\infty}(y\mu({\{x: f(x) > y\}}))=0$.

So all we're left with in the end is $\int_0^\infty f^*(y)dy$, which is the definition in the Wikipedia article. So at least that makes sense to my satisfaction. I still don't understand, however, Wikipedia's intuitive justification for the result, specifically why the area of the infinitesimal horizontal strip between $y=t$ and $y=t+dt$ is $\mu({\{x: f(x) > t\}})dt$.

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Forgive my MS Paint, and let $dt>0$ be a tiny change in height. I believe the Wikipedia argument can be intuited as follows. We approximate the integral of $\color{blue}f$ with the area of the $\color{red}{\text{rectangles}}$. enter image description here

$\color{green}{\text{(1)}} = \{x : f(x) > 0\} = [f>0]$. (This is our entire support.) The area of the first rectangle is $\mu(f>0)dt$.

$\color{green}{\text{(2)}} = [f>dt]$. The area of the second collection of rectangles is $\mu(f>dt)dt$

In each case, note that the base of each rectangle at height $t$ is $[f>t]$. In the end we are left with $$\int f \approx \sum_{k=1}^\infty \mu(f>k dt)dt$$ and in the limit $dt\to 0$ we should expect to recover the form as claimed, $\int f = \int_0^\infty \mu(f>t)dt$.

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