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Hi: I'm reading some notes on measure theory and I don't understand one of the steps where the author defines the integral of f with respect to $\mu$.

The link is here: http://www.martinorr.name/2008/probability/PM.pdf and the paragraph i am referring to is on page 11. I will repeat it here for convenience.

$\int \left(\sum_{j=1}^{n} \beta_{j}f_{j} \right) du = \sum_{j=1}^{n} \beta_{j} \int f_{j} du $ ( where the $f_{j}$ are simple functions. )

The simple functions are good enough to approximate all measurable functions from below. given $f \in mS^{+}$, let

$ f_{n}(x) = 2^{-n}\left[2^{n} \times min(f(x), n)\right] = 2^{-n} \sum_{j=1}^{n2^{n}} I_\left({x:f(x) \ge j2^{-n}}\right)$

Then $f_{n}$ is an increasing sequence which converges to f pointwise and ineeded $f_{n} \rightarrow f$ uniformly on any set of the form $ (x:f(x) \le k)$ .

I don't see how $f(n)$ accomplishes this ( the immediately previous sentence ) ? I plugged in $n =1$ and got two terms which were $I(x: f(x) \ge \frac{1}{2})$ and $I(x: f(x) \ge 1)$.

Then I did the same thing for $n=2$ and got 8 terms that ran from $\frac{1}{4}, \frac{2}{4} \ldots , 8 $. For $n = 3$, one obtains 24 terms.

But doing this is not helping me see what is happening. Does any know of a picture that shows this ? I think that understanding this may be critical to understanding the difference between lebesgue integration versus reimann integration so if anyone has any insights or references or explanations, it's much appreciated.

mark leeds
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  • The author could have made it a little simpler by using the equivalent $f_n = \sum_{j=1}^{n 2^n} j 2^{-n} 1_{f^{-1}([j 2^{-n}, (j+1)2^{-n}))} + n 1_{f^{-1}([n,\infty))}$. This takes slices of the range up to $n$ (and stuffs $[n,\infty)$ into one 'bucket') instead of the sort of 'stacked pie' approach taken by the author. Unless I made a mistake, the two definitions are equivalent, this one has the advantage of being easier to understand (I think) and the sets underlying the simple funcions are disjoint. – copper.hat Feb 28 '16 at 22:40
  • The key to Lebesgue integration is that the domain can be sliced in many more ways than with Riemann integration (only partitions). – copper.hat Feb 28 '16 at 22:48

1 Answers1

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Consider a nonnegative function $f$ with the given approximating simple functions $f_n$. If $f(x) \leq n$, then $|f_n(x)-f(x)| \leq 2^{-n}$; if this were not true, then $f_n(x)$ would be bigger or smaller, as follows from its definition. If $f$ is a real valued function (i.e. it never takes the value $+\infty$) then for all sufficiently large $n$ you have $f(x) \leq n$. So $|f_n(x)-f(x)| \leq 2^{-n}$ for all sufficiently large $n$, which does the job.

Visually, you are drawing horizontal lines $y=0,y=2^{-n},y=2\cdot 2^{-n},y=3 \cdot 2^{-n},\dots,y=n,y=+\infty$, then you are grouping all the points where the graph of $f$ is between two adjacent horizontal lines. You then take a simple function which takes on the lower value in each of these subsets. By contrast Riemann integration effectively draws vertical lines, but is otherwise analogous.

Ian
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  • thanks. what both of you said sounds very clear I have to print out read it very carefully to see if I "get it". but if anyone knows of a picture that describes what is going on, that would probably help me a lot. all the best and thank you very much copper.hat and ian. – mark leeds Feb 28 '16 at 23:42
  • Hi All: I read the explanations and I understand copper.hat's equivalent function but I think I'm missing something. $f(x)$ returns a value of $y$ for a given $x$. So how could a function $f_{n}$ which is composed of indicator functions ever approach $f$ since it's returning $1 \times j \times 2^{-n}$ whenever the set satisfies that condition of the indicator function. thanks. – mark leeds Feb 28 '16 at 23:55
  • @markleeds Let's do an example. Suppose $f(x_0)=1.1$. Now $f_1(x)$ is one of $0,1/2,1$. $f_1(x_0)$ will be the closest of these values to $f(x)$ i.e. $1$. Next $f_2(x)$ is one of $0,1/4,2/4,3/4,1,5/4,6/4,7/4,2$. $f_2(x_0)$ is the closest of these to $f(x_0)$, which is again $1$. Next $f_3(x)$ is one of $0,1/8,2/8,\dots,3$. $f_3(x_0)$ is the closest of these to $f(x_0)$, which is $9/8=1.125$. And so forth. Thus once $n \geq f(x)$, you have $|f_n(x)-f(x)| \leq 2^{-n}$, which gives the pointwise convergence. – Ian Feb 29 '16 at 00:00
  • thanks so much ian. I think I get it now. I appreciate your patience and help. – mark leeds Feb 29 '16 at 02:02