10
$\begingroup$

I know the definition, from A.N. Kolmogorov and S.V. Fomin's Элементы теории функций и функционального анализа, of Lebesgue integral of measurable function $f:X\to \mathbb{C}$ on $X,\mu(X)<\infty$ as the limit$$\int_X fd\mu:=\lim_{n\to\infty}\int_Xf_nd\mu=\lim_{n\to\infty}\sum_{k=1}^\infty y_{n,k}\mu(A_{n,k})$$where $\{f_n\}$ is a sequence of simple, i.e. taking countably many (not necessarily finitely) values $y_{n,k}$ for $k=1,2,\ldots$, functions $f_n:X\to\mathbb{C}$ uniformly converging to $f$, and $\{y_{n,k}\}=f_n(A_{n,k})$ where $\forall i\ne j\quad A_{n,i}\cap A_{n,j}=\emptyset$. If $\mu$ is not finite but is $\sigma$-finite with $X=\bigcup_{j=1}^\infty X_j$, $X_1\subset X_2\subset\ldots$ then $\int_{X}fd\mu:=\lim_j\int_{X_j}fd\mu$ if such limit exists for any sequence $\{X_j\}$ (said to be exhaustive) such that $X=\bigcup_{j=1}^\infty X_j$, $X_1\subset X_2\subset\ldots$.

I read other authors defining the Lebesgue integral for non-negative functions $g:X\to[0,+\infty]$, where $\mu(X)\le\infty$, by using simple non-negative functions $s_i$ taking only finitely many values on measurable sets $A_i$ in the following way:$$\int_X gd\mu:=\sup\Bigg\{\sum_i s_i\mu(A_i):\forall x\in X\quad s_i(x)\le g(x)\Bigg\}$$ and in general in the following way, provided that $\int_X|f|d\mu<\infty$:$$\int_Xfd\mu:=\int_X \text{Re}f_+d\mu-\int_X \text{Re}f_-d\mu+i\Bigg(\int_X \text{Im}f_+d\mu-\int_X \text{Im}f_-d\mu\Bigg)$$ where subscript $\pm$ is used to define $g_+(x):=\max\{g(x),0\}$ and $g_-(x):=-\min\{g(x),0\}$ for any measurable function $g$.

I have not been able to find more about it, since the texts that are available to me all use Kolmogorov-Fomin's definition, but I am convinced that such definitions are equivalent.

From what I know -I have finished Kolmogorov-Fomin's Элементы теории функций и функционального анализа, which significantly overlaps with Introductory Real Analysis by the same authors- I would say that it would be enough to prove the equivalence for non-negative functions $g:X\to[0,+\infty)$.

If they are equivalent, how can it be proved? I have not been able to use what I know, Kolmogorov-Fomin's content, to prove it: is a much more advanced knowledge necessary? I have tried by using the density in $L^1(X)$ of the set of measurabe functions taking finitely many values on finite measure sets, and also the fact that, if $\mu(X_j)<\infty$ and $\{s_k\}$, with $s_k:X_j\to\mathbb{C}$, converges to $g:X_j\to\mathbb{C}$ in the metric of $L^1(X_j)$, then a subsequence of $\{s_k\}$ uniformly converges to $g$, but I have not been able to prove the equivalence of the two definitions. I uncountably thank you!

$\endgroup$
  • 1
    $\begingroup$ My copy of Kolmogorov's book (English translation) says at most countably many $y_k$, and when summing he never sums up to $\infty$ leaving it open (so you can have a finite or infinite sum). $\endgroup$ – Aram Nov 12 '14 at 18:03
  • $\begingroup$ @Aram: yes, mine does too. By "countably many" I intended a finite number or a set of values of cardinality $\aleph_0$. Thank you very much! $\endgroup$ – Self-teaching worker Nov 12 '14 at 22:40
  • 1
    $\begingroup$ You might want to check Royden's Real analysis book. $\endgroup$ – Batman Jan 7 '15 at 19:51
  • $\begingroup$ @Batman Thank you very much! It appears to use the second definition I wrote -as possibly most more recent sources do-, but I can find nothing explicitly proving the equivalence with Kolmogorov-Fomin's definition... $\endgroup$ – Self-teaching worker Jan 7 '15 at 22:08
  • 2
    $\begingroup$ Whichever definition you take: after you develop enough properties of the integral, then you should be able to easily show the equivalence with the other definition. If it does not seem easy to you, then you have not developed enough of the properties! $\endgroup$ – GEdgar Jan 7 '15 at 22:45
2
+50
$\begingroup$

As was mentioned in comments by GEdgar, it can be derived from the basic properties obtained from both definitions (basic properties of the "unusually" defined integral could be found, for instance, in Bogachev-Smolyanov textbook). By standard convergence theorems (Lebesgue, Levi) the problem could be reduced to comparing integrals of functions taking finitely many values, which in turn reduce by linearity to the indicator functions of measurable sets and the latter case is obvious from the definition. I will give a direct proof though.

Denote by $I_1(f)$ the integral in the first definition and $I_2(f)$ the one in the second. Assume that $0\le f<\infty$ and $\mu(X)<\infty$. Take $y_{n,j}=j/n$ and $A_{n,j}=f^{-1}[j/n,(j+1)/n)$. Since $I_1(f)=\lim\limits_{n\to\infty}\lim\limits_{k\to\infty}\sum\limits_{j=0}^k y_{n,j}\mu(A_{n,j})$ and each $\sum\limits_{j=0}^k y_{n,j}I_{A_{n,j}}\le f$, it follows that $I_1(f)\le I_2(f)$.

To prove the reverse inequality, take $y_{n,j}=j/n$ and $A_{n,j}=f^{-1}((j-1)/n,j/n]$. Then, for every $s(x)=\sum_i s_i\mu(B_i)$ such that $s\le f$, we have $$ \sum s_i\mu(B_i)= \sum_{i,j} s_i\mu(B_i\cap A_{n,j})\le \sum_{i,j} y_{n,j}\mu(B_i\cap A_{n,j})\le \sum_{j} y_{n,j}\mu(A_{n,j})\to I_1(f). $$ Thus, since $s$ was arbitrary, $I_2(f)\le I_1(f)$.

If $X=\bigcup X_i$, $\mu(X_i)<\infty$, by definition $$ I_1(f)=\lim I_1(f I_{X_i})=\lim I_2(f I_{X_i})\le I_2(f). $$

For every simple function $s\le f$, $$ I_2(s)=I_1(s)=\lim I_1(s I_{X_i})\le\lim I_1(f I_{X_i})\le I_1(f), $$ so $I_2(f)\le I_1(f)$. The monotone property of $I_1$ and $I_2$ for positive functions can be seen directly from definitions.

Finally, if $\mu(f=\infty)>0$, both integrals can be checked to be infinite by choosing some approximating functions including $\infty\cdot I_{(f=\infty)}$ term. The case of signed/complex $f$ follows immediately, since they are defined using positive/negative parts.

$\endgroup$
  • $\begingroup$ Detailed and clear answer, so clear that even I can understand. Thank you a lot! $\endgroup$ – Self-teaching worker Jan 14 '15 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.