Differences

This shows you the differences between two versions of the page.

--- math105-s22:s:rasmuspallisgaard:start [2022/02/28 22:35]
pallisgaard description
+++ math105-s22:s:rasmuspallisgaard:start [2026/02/21 14:41] (current)
@@ Line 41: / Line 41: @@
 {{ :math105-s22:s:rasmuspallisgaard:maht105-hw5-rasmus-pallisgaard-2.pdf |Homework 5}}
+{{ :math105-s22:s:rasmuspallisgaard:analysis_ii_-_notes_3.pdf |Homework 6}} (There is a mistake in problem 1, b. I'll fix it and update the file.)
+**Resume of Lebesque measure theory**
+*This is gonna be a rough summarisation of all we've covered in Lebesque measure theory so far. It will probably not contain everything of importance and might have some gaps that I didn't think to cover. If you find gaps like these, please do write to me on Discord so I can review these. Thanks!*
+Our venture into Lebesque measure theory begins with the definition of the outer measure - a measure of a subset $A$ found by
+$$
+m^*(A)=\inf\left\{ \sum_k|B_k|:\{B_k\} \text{ is a covering of } A \text{ by open boxes} \right\}
+$$
+Useful from this definition and this section is the definition that sets with outer measure zero is called a zero set. Boxes are created from intervals $(a_i,b_i)$ since the measure of an interval is its end point minus its starting point. The proofs of useful properties, such as monotonicity, sub-additivity, etc. are most often proved using the $\epsilon$ trick.
+A set $A\subset R$ is then measurable if it the division $A|A^c$ is so *clean* that for all subsets $X\subset R$,
+$$
+m^*(A)=m^*(X\cap A) + m^*(A\cap E^c)
+$$
+Although I personally like the Tao condition better:
+$$
+m^*(A)=m^*(X\cap A) + m^*(A\setminus E)
+$$
+Here we see that by this definition, additivity is achieved. Throughout the course we have proved a lot of properties of measurable sets, including sub-additivities of outer measures, monotonicity, etc. As mentioned before, for this we often use the $\epsilon$ trick - a trick that is also used to prove measurability of a closed interval, a zero set, and a closed box in n dimensions.
+We then went on to proof that the Lebesque measure is **regular** in the sense that a measurable set $E$ can be sandwiched between an $F_\sigma$-set and a $G_\delta$-set such that $F_\sigma\subset E\subset G_\delta$. Here an $F_\sigma$-set is a countable union of closed sets $F_\sigma=\cup^\infty_iF_i$ and a $G_\delta$-set is a countable intersection of open sets $G_\delta=\cup^\infty_iG_i$.
+The proof of this uses the fact that can define a decreasing sequence of open sets from $\mathbb{R}$ such that the measure of these set sequences goes to the measure of $E$. You can then define a closed increasing sequence from the complement of one of these sequences. The major step here is then to show that this complement set has the same measure as $E$.
+We then covered the theorems of products and slices. The theorem of measurable products says that if sets $A$ and $B$ are measurable, then $m(A\times B)=m(A)\cdot m(B)$. The proof of this uses hulls and kernels of measurable sets (These are $F_\sigma$-sets and $G_\delta$-sets).
+Afterwards we proved that if $E$ is measurable then it has measure zero iff almost every slice of $E$ has measure zero. A slice $E_x$ of a set $E\subset R^n\times R^k$ is defined as $E_x=\{y\in R^n:(x,y)\in E\}$.
+The prove of the above theorem is bit more involved, but essentially boils down to first finding that the set $E$ has the same measure as the set $E$ with all nonzero slices removed. Afterwards one seeks to prove that since all these slices has measure zero, then measure of $E$ is zero. We then, for a slice of any compact $K\subset E$, surround it by a a long but thin compact box $W(x)$. Since these boxes are thin, but not zero width, we can cover the set by a countable amount of these boxes. By disjointizing the widths we can find that the boxes have measure zero, so measure of $K$ is zero. Then inner measure is zero (see definition of inner measure with respect to closed subsets) is zero, and by measurability measure of $E$ is zero.
+Proving the other direction, if $E$ has measure zero, then there exists a $G_\delta$-set $G\supset E$, The main step is to set up $X(\alpha)=\{x:m(G_x)>\alpha\}$ for each slice $G_x$. One can then finalise the proof using the same disjointizing method on a compact set $K(x)$ contained in $G_x$ with $m(K(x))=m(G_x)$ and a neighbourhood $W(x)$.
+{{ :math105-s22:s:rasmuspallisgaard:math105-hw5-rasmus-pallisgaard.pdf |Homework 5}}
+{{ :math105-s22:s:rasmuspallisgaard:math105_hw6_rasmus_pallisgaard.pdf |Homework 6}}
+{{ :math105-s22:s:rasmuspallisgaard:math105-hw7-rasmus-pallisgaard.pdf |Homework 7}}
+{{ :math105-s22:s:rasmuspallisgaard:math105_hw8.pdf |Homework 8}}
+{{ :math105-s22:s:rasmuspallisgaard:math105_hw9.pdf |Homework 9}}
+{{ :math105-s22:s:rasmuspallisgaard:math105_hw10.pdf |Homework 10}}
+{{ :math105-s22:s:rasmuspallisgaard:math105_hw11.pdf |Homework 11}}
+{{ :math105-s22:s:rasmuspallisgaard:math105-hw12-rasmus-pallisgaard.pdf |Homework 12}}
+====Final Essay====
+Here is my final essay on lebesque integration and measure theory, and why its needed and relevant in the context of integration.
+{{ :math105-s22:s:rasmuspallisgaard:final_essay.pdf |Why Do We Need Measure Theory?}}

Lecture Notes

User Tools

Site Tools

Differences

Page Tools