Lebesgue integration

**southprkfan1** · April 14th 2010, 04:56 PM

Let f be a (Lebesgue) integrable function and A be a measurable set.

Show that:

$\int lfl\chi_A(x) \leq K*m(A)$ for some K

I know that since f is integrable we can say $\int lfl \leq K$ , and this feels like it should be a two line proof, but I can explicitly show how to get to the next step.

**maddas** · April 15th 2010, 07:04 AM

Disregard the following... (facepalm)

Let s be a simple function, $0\le s\le |f|1_A$ so $s=s1_A$ . Call such a function "good". We aim to show the integral of any good function is bounded by $K\cdot m(A)$ so $\int |f|=\sup\{\int s : \text{s good}\}$ is too.

Since |f| is integrable, the set $E:=\{\int t : 0\le t \le |f|, \text{t simple}\}$ is bounded above and we can chuse a simple function $0\le t' \le |f|$ whose integral is close to the supremum. Then $t := t' + k$ is simple and greater than every function in E for large k (oopsthis part was wrong; thanks Opalg). Suppose s is a good function. Then s is in E, so t is greater than it, and it follows that t1_A is greater than it. So $t \ge s$ for all good s and it follows that $\int t \ge \int s$ for all good s so that $\int t \ge \int |f|$ . Now since t is simple, $t = \sum_{i=1}^N t_i 1_{B_i}$ , so $\int t1_A = \sum t_i m(B_i\cap A) \le (\max t_i) \sum m(B_i\cap A) \le (\max t_i)\;Nm(A)$ and we are done.

(Note that I've abused notation by writing E for both the set of all simple functions between 0 and |f|, and the integrals of such functions. I hope its not confusing.)

**Opalg** · April 15th 2010, 11:18 AM

Originally Posted by southprkfan1

Let f be a (Lebesgue) integrable function and A be a measurable set.

Show that:

$\int lfl\chi_A(x) \leq K*m(A)$ for some K

I know that since f is integrable we can say $\int lfl \leq K$ , and this feels like it should be a two line proof, but I can explicitly show how to get to the next step.

Assuming that you want K to be a constant independent of A, this result looks wrong.

Take f to be the function $f(x) = x^{-1/2}$ on the interval (0,1). If A is the interval (0,1/n) then $\int\!|f|\chi_A^{\mathstrut} = \int_0^{1/n}\!\!\!\! x^{-1/2}dx = \Bigl[2x^{1/2}\Bigr]_0^{1/n} = \frac2{\sqrt n}$ . If $\int\!|f|\chi_A^{\mathstrut} \leqslant Km(A)$ then $\frac2{\sqrt n} \leqslant\frac Kn$ , or $\sqrt n\leqslant K/2$ . That clearly cannot hold for all n.

**mabruka** · April 15th 2010, 12:00 PM

I think $K$ does not depend on the set $A$, but it depends on the function f.

Since f is integrable then it is bounded almost everywhere and using integral monotonicity we get the desired conclusion.

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