3
$\begingroup$

Please how can I justify that these two definitions can be the same thing or let me say how can I compare the following two definitions.

  1. If $g$ is a right continuous, increasing step function on $\mathbb{R}$ with countable discontinuous points $x_1,x_2,\cdots,x_n$. Then for any function $f$, we define the integral $$\int_{\mathbb{R} }f dg =\sum_{k=1}^{n} f(x_k)\Delta g(x_k)$$where $\Delta g(x_n) = g(x_n^+)-g(x_n^-)> 0$.

  2. Let $(X, \mathcal{H}, \mu)$ be a measure space and let $f: X \rightarrow [0, \infty]$ be a measurable function. We define the Lebesgue integral of $f$ over $X$ as $$\int_{X}f d \mu= \sup \left\{ \int_{X}g d \mu: 0 \le g \le f, g \,\, \text{is simple} \right\}.$$

I would be glad if a proof or correspondence can be given. Thanks for your help.

$\endgroup$
  • $\begingroup$ Try to define $g:\mathbb{R}\to\mathbb{R}$ by $g(x)=\mu(-\infty, x]$ and see what happens with $\int_{\mathbb{R}} fdg$. $\endgroup$ – Filburt Feb 19 '17 at 23:40
  • $\begingroup$ As a physicist by education, I've never found any use for Lebesgue measures and its associated integrals. I have stuck with the Riemann interpretation and find that an integral is an integral is an integral. That may be simple minded, but it has always worked for the universe I'm living in. $\endgroup$ – Han de Bruijn Feb 24 '17 at 11:56
1
$\begingroup$

The second definition contains the first:

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ as in 1.. With the classical extensions theorems, one can construct a measure $\mu_{g}$ over the borelians such that $\mu_{g}(]a,b])=g(b)-g(a)$. It can be show that $\mu_{g}(A)=\sum_{x_{n}\in A}\Delta g(x_{n})$.

  • If $f=\sum_{i=1}^{m}a_{i}1_{A_{i}}$ is measurable (and the $A_{i}$ disjoint) then $$\int fd\mu_{g}=\sum_{i=1}^{m}a_{i}\mu_{g}(A_{i})=\sum_{i=1}^{m}a_{i}\sum_{x_{n}\in A_{i}}\Delta g(x_{n})=\sum_{i=1}^{m}\sum_{x_{n}\in A_{i}}f(x_{n})\Delta g(x_{n})=\sum_{x_n}f(x_{n})\Delta g(x_{n})$$

  • If$f$ is positive and measurable, then there exist simple functions $f_{k}$ such that $f_{k}(x)\rightarrow f(x)$ monotonously. We deduce, by the monotone convergence theorem and the previous result, that $\int fd\mu_{g}=\sum_{n}f(x_{n})\Delta g(x_{n})$

$\endgroup$

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.