Cezar Port

Infinite valued simple measurable functions

December 13, 2018December 14, 2018Cezar PortLeave a comment

Definition. Let $( X, \Sigma, \mu )$ be a measure space. A measurable $(\Sigma \ | \ \mathcal{\overline {B}} )$ (the extended borel sets of the extended reals $\mathbb{\overline{R}}=[-\infty, \ \infty ]$ ) function that is nonnegative and it’s range is a finite set is named a simple nonegative measurable function. $L^+_0$ is the set that represents all this type of functions.
Proposition. Let $( X, \Sigma, \mu )$ be a measure space. $\{ f_n \}_{n\in \mathbb{N}} \in L^+_0$ is an increasing sequence of functions. While $g \in L^+_0$ is such that $\lim_{n \to \infty} f_n(x) \geq g(x)$ , for all $x \in X$ .
Then $\lim_{n\to\infty} \int_{X} f_n d\mu \geq \int_{X} g d\mu$ .
Literature Review.
1. This is a generalized version of Theorem 4.2.1 from the book “A basic course in Measure and Probability” by Ross Leadbetter, Stamatis Cambanis, Vladas Pipiras. Only that in the book, the simple nonnegative measurable functions are not allowed to take $\infty$ value. Which is also the main reason of this post- in order to generalize the theorem. But the above mentioned authors do not give a reason as to why, $\infty$ is not allowed. A reason could be that it is easier to do develop the theory, but at the cost of having an unexplained detail left out, which ruins right from the beginning the development of the theory. Though the pedantic thinker easily lifts this annoyance of his chest, once it is noticed that in the definition for the nonnegative measurable function ( and nonnegative function have the right of having the infinity value, as they are and seen as limits of nonnegative simple functions)
$\int_X f=sup\{ \int_X s \ |\ 0\leq s\leq f, s \in L^+_0 \}= sup\{ \int_X s \ |\ 0\leq s<\infty,\ 0\leq s\leq f, s \in L^+_0 \}$ .
2. In Inder K. Rana book “An introduction to Measure and Integration“, the approach is like mine’s, the simple functions are allowed to take $\infty$ value, (Definition 5.1.1, page 118). But there is afterwards a mistake in proving, the “almost” same proposition, as ours,(Proposition 5.1.10, page 121). The mistake is forgetting that the simple nonnegative function can have the $\infty$ value. Basically, there is considered the set, for a fixed $0<c<1$ , $B_n=\{x | g\geq cs\}$ (in the condition there is taken for granted that $s=\lim f_n , \ s\in L^+_0$ ), which would next imply that $\lim B_n=X$ . Which is wrong, as $B_n$ does not take into account that $s$ can also take $\infty$ .
3. Most of the books, though, consider the simple functions to assume only real values. Examples: J.Yeh, “Real analysis, Theory of Measure and Integration.“, Halmos’s Book (at page 84-85), Gerard Folland “Real analysis, modern techniques and their applications“(page 46), Heinz Bauer “Measure and Integration Theory“, and in general mostly everywhere.

Solution 1. Let $G_1=g^{-1}(\infty)$ . $G_1$ might just as well be $\emptyset$ . And let $G_2=g^{-1}(g_2), \dots G_k=g^{-1}(g_k)$ where $g(X\setminus G_1)=\{g_2,..., g_k\}$ , and $i\neq j \implies g_i\neq g_k$ .
Fix $m \in \mathbb{N}, m\geq 1$ and consider the set
$S_{n,m}=\{ \ x\in G_1 \ | f_n(x)\geq m \} \cup \{x\in G_2\cup ... \cup G_k \ | \ f_n(x)\geq (1-\frac{1}{m})g(x) \ \}$ .
The following hold, since $f_n$ , are increasing $\lim_{n \to \infty } S_{n,m}\cap G_i =G_i$ for each $i =\overline{1,k}$ . Then $\int_{X} f_n d\mu \geq m\mu(S_{n,m}\cap G_1) +(1-\frac{1}{m})g_2\mu(S_{n,m}\cap G_2)+...+(1-\frac{1}{m})g_k\mu(S_{n,m}\cap G_k)$ . Taking the limit wrt $n$ , we get $\lim \int_{X} f_n d\mu \geq m\mu(G_1) +(1-\frac{1}{m})g_2\mu(G_2)+...+(1-\frac{1}{m})g_k\mu(G_k)$ . Now, taking the limit wrt $m$ , $\lim \int_{X} f_n d\mu \geq \infty\mu(G_1) +g_2\mu(G_2)+...+g_k\mu(G_k) = \int_X g d\mu$ .
Solution 2. Once the theory is developed through the real valued simple functions, one can consider $f=\lim f_n$ , and see $f,g$ as measurable functions. Because $f\geq g$ , then $\lim_{n \to \infty} \int_X f_n d\mu=\int f_X d\mu \geq \int g_X d\mu$ .

Note: The convention is $\infty 0=0 \infty=0$ . There are some good reasons for this, but it is not a topic to be explained here. But here is an argument: in order for some “limits” to behave good. It is important to realize that all these “taking of limits” is possible only because the sequence under which the limit is taken is an increasing sequence. For example: $\lim_{n \to \infty} \infty(\frac{1}{n})=\infty\neq 0 =\infty 0 = \infty*\lim_{n \to \infty} \frac{1}{n}$ . In our case this situation does not happen as if $\lim_{n \to \infty} c_n=0, \ \ \{c_n\}$ is an increasing sequence of numbers from $[0, \infty]$ , then all $c_n=0$ , and $\lim \infty c_n=0=\infty \lim c_n$ .

Teorema lui Sondat

September 20, 2016September 24, 2016Cezar PortLeave a comment

Teorema lui Sondat

v In geometrie sunt doua teoreme atribuite lui Sondat. Una spune ca axa de perspectivitate, a doua triunghiuri paralogice, taie in jumate segmentul dat de ortocentrele triunghiurilor. Iar a doua ca daca 2 triunghiuri sunt ortologice si perspective atunci centru de perspectivitate si cele doua centre ortologice sunt colineare pe o dreapta perpendiculara axei de perspectivitate.

vAici prezint o solutie sintetica la a doua teorema a lui Sondat.

Solutie. Fie ${\triangle ABC}$ si ${\triangle X_AX_BX_C}$ cele doua triunghiuri ortologice si perspective, cu ${X}$ un centru ortologic astfel incit ${XB \perp X_AX_C, XC\perp X_AX_B}$ si ${XA\perp X_BX_C}$ . Din definitie punctele ${P_A = X_BX_C\cap BC}$ , ${P_B = X_AX_C\cap AC}$ si ${P_C = X_BX_A\cap BA}$ sunt colineare. Iar ${P=XX_A\cap XX_B\cap XX_C}$ e centrul de perspectivitate.

v ${h_c}$ e omotetia, cu centru ${C}$ , ce duce ${X_C}$ in ${P}$ . ${h_b}$ e omotetia, cu centru ${B}$ , ce duce ${X_B}$ in ${P}$ . Deci ${h_b(X_BX_C)=h_c(X_BX_C)}$ , ce implica ca ${h_b(P_A)=h_c(P_A)=Q_A}$ . Prin umare, ${h_b(P_B)=h_c(P_B)=Q_B}$ si ${h_b(P_C)=h_c(P_C)=Q_C}$ . Punctele ${Q_A,Q_B,Q_C}$ satisfac relatiile: ${Q_AP \perp AX}$ ; ${Q_B \perp BX}$ ; ${Q_CP \perp XC}$ . ${\bigstar}$

Lema. Fie punctele ${A,B,C,P,Q}$ si ${\Gamma}$ , conicul ce le contine. ${Q_A,Q_B,Q_C}$ sunt intersectiile perpendicularelor din ${P}$ pe ${AQ,BQ,QC}$ cu ${BC, AC,AB}$ , respectiv. ${Q_A,Q_B,Q_C}$ sunt colineare ${\iff}$ ${\Gamma}$ e o hiperbola dreptunghica.

vDemonstrare. Fie ${Q_A,Q_B,Q_C}$ colineare si ${P_A,P_B,P_C}$ celelalte intersectii ale dreptelor ${Q_AP,Q_BP,Q_CP}$ cu ${\Gamma}$ . Teorema lui Pascal pentru ${ABCPP_AP_B}$ da ${Z=AP_A \cap BP_B \cap Q_AQ_B}$ . Similar ${Z'=AP_A \cap CP_C \cap Q_AQ_B}$ . Prin urmare ${Z=Z'}$ si ${Z=Q_AQ_B \cap BP_B\cap CP_C\cap AP_A}$ .

vDin Teorema lui Pascal pentru ${ABP_BP_APQ}$ rezulta ca ${Z}$ , ${PP_B\cap AQ=M}$ si ${PP_A \cap BQ=N}$ sunt colineare. La fel ${Z}$ , ${PP_C\cap AQ=M'}$ si ${PP_A \cap CQ=N'}$ sunt colineare. Insa ${P}$ e ortocentru in ${\triangle M'N'Q}$ si ${\triangle MNQ}$ . Deci ${M'N'\parallel MN}$ ${\Longrightarrow}$ ${Z}$ e un punct de la infinit.

vAsadar, ${AP_A \parallel BP_B \parallel MN \perp PQ \perp Q_AQ_BQ_C}$ . Atunci ${P}$ e ortocentru ${\triangle AP_AQ}$ . Concluzionind ca ${\Gamma}$ e o hiperbola dreptunghica. Conversa se face prin acelasi rationament. ${\square}$
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vDoarece ${Q_AQ_BQ_C \parallel P_AP_BP_C}$ , ${\bigstar}$ si lema obtinem ca ${PX \perp P_AP_BP_C}$ . Similar ${X}$ si celalalt centru de ortologie se afla pe o linie perpendiculara liniei ${P_AP_BP_C}$ , concluzionind solutia.
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A fundamental property of Circular Cubics

September 19, 2016September 19, 2016Cezar PortLeave a comment

v Circular cubics are cubics which contain the circular points at infinity. For example, the Neuberg Cubic is circular. Any self-isogonal cubic wrt ${\triangle ABC}$ , with pivot a point on the line at infinity, is circular. Inverting a conic wrt to a point on the conic gives a Circular Cubic.Which would be an acnode, a cusp or a crunode, in case of inverting an ellipse, a parabola or a hyperbola. Clearly the inversion point is the double point of the cubic.

vAny conic passing through any four fixed points of a cubic intersects it in two other points, which define the line ${l}$ . It’s well-known that ${l}$ passes through a fixed point from the cubic, as the conic varies, but when the cubic is circular there is a special case.

Property. Let a circle intersect the circular cubic in four points. Then the line joining the points given by the intersection of any opposite lines of the quadrangle with the cubic, is parallel to the cubic’s real asymptote.

vHere I present a sythetic solution to this property only when the cubic is the inverse of a conic.

Solution. Let ${X}$ be the double point of the circular cubic. By inversion in ${X}$ , the cubic will transform into the conic ${\Gamma}$ ; the circle in another circle intersecting ${\Gamma}$ in ${A,B,C}$ and ${D}$ ; the opposite lines in ${\odot XAC}$ and ${ \odot XBD}$ intersecting the conic in ${E,F}$ , respectively. Clearly ${E,F}$ are the the intersection of the opposite lines of the quadrangle with the cubic. Therefore it suffices to prove that ${\odot XEF}$ is tangent to ${\Gamma }$ in ${X}$ .

Lemma. The line joining the two points of intersection of a circle containing two fixed points of a fixed conic ${\Omega}$ , with ${\Omega}$ passes through a fixed point from the line at infinity.

Proof. Let ${\odot \alpha}$ and ${\odot \beta }$ with ${A, B}$ common points. ${C,D \in \odot \alpha }$ . Let the conic ${\Omega}$ passing through ${A,B,C,D}$ intersect ${\beta }$ in ${E,E'}$ . It suffices to prove ${EE' \parallel CD}$ .

v ${{M,M'} = {AD,CB}\cap \beta}$ and ${{N,N'} = {DB,AC}\cap \beta}$ . By Reim’s Theorem, ${CD \parallel MM' \parallel NN'}$ ${\bigstar}$ . Since ${A,B,C,D,E,E' \in \Omega}$ , results ${A(D,C;E,E')=B(D,C;E,E')}$ or ${A(M,N'E,E')=B(N,M';E,E')}$ .Combining this with ${\bigstar}$ and considering that ${\beta}$ is a circle, gives that ${EE' \parallel CD}$ .

vSo by the lemma ${XE\parallel BD}$ and ${XF \parallel AC}$ . Now ${\odot XEF}$ can’t intersect ${\Gamma}$ in another point ${X'}$ . Cause then by the lemma ${X'F \parallel AC \parallel XF}$ , which is impossible. Therefore ${\odot XEF}$ is tangent to ${\Gamma}$ .
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The Neuberg Cubic

September 9, 2016September 19, 2016Cezar PortLeave a comment

This page is devoted to synthetic solutions to different properties of the Neuberg Cubic of a triangle.

v The Neuberg Cubic is a pivotal isogonal cubic, with pivot point $X\left(30\right)$ =Euler infinity point. In other words it is the locus of points $P$ for which $P$ , it’s isogonal conjugate and $X\left(30\right)$ are collinear.

Property 1.Let $P_A,P_B$ and $P_c$ be the reflections of $P$ in the sidelines of $\triangle ABC$ . Let $\triangle P_AP_BP_C$ be called the $P$ -reflection triangle wrt $\triangle ABC$ . $P\in \rm{Neuberg \ Cubic} \Longleftrightarrow$ $P$ -reflection triangle is perspective with $\triangle ABC$ .

Proof.

Notations:

For a point $X$ ,

$X_A$ is the reflection of $X$ in $CB$ .( $X_B$ and $X_C$ are defined similarly)
line $x_A=X_BX_C$ . ( $x_B,x_C$ are defined similarly )

For a line $l$ , $A^l$ is it’s intersection with $BC$ .( $B^l,C^l$ are defined similarly)

Lemma 1. Let $\triangle ABC$ and $X \in BC$ . The the locus of points $P$ such that $X\in p_A$ is a circle passing through $X_B,X_C$ and $A$ .

Proof. Let $l$ be any line containing $X$ . The reflection of $l$ in $AB$ , $l_{rc}$ , passes through $X_C$ . $l_{rb}$ is defined the same way and it passes through $X_B$ .
vvLet $l_{rb}\cap l_{rc}=L$ , then $l_A=l$ . When $l$ moves clockwise around $X$ then $l_{rc}$ moves counterclockwise around $X_C$ and $l_{rb}$ moves counterclockwise around $X_B$ . Because they all rotate with the same angle, it follows that $L$ belongs to a circle, which contains the points $X_B,X_C$ and clearly $A$ (when $l=XA$ ).

nb2

Lemma 2. Let $\triangle ABC$ and line $l\perp \rm{Euler \ line}$ . The triangle given by the lines $B^{l}_{A}A^{l}_{B}$ , $A^{l}_CC^{l}_{A}$ and $C^{l}_BB^{l}_{C}$ is homothetic with $\triangle DEF$ , the pedal triangle of $H$ wrt $\triangle ABC$ ,with homothety center $H$ .

Proof. The homothety, wrt $C$ , which transforms $l$ into the orthic axis, sends $B^l_AA^l_B$ to $DE$ . So $B^l_AA^l_B \parallel$ $DE$ . Similarly $A^l_CC^l_A \parallel$ $DF$ and $C^l_BB^l_C \parallel$ $EF$ .
v $Q$ is the projection of $A^l$ on $FE$ . Then $A^l_CQ_C$ is the distance between $FD$ and $A^l_CC^l_A$ . Also $A^l_BQ_B$ is the distance between $DE$ and $A^l_BB^l_A$ . Clearly $A^l_CQ_C = A^l_BQ_B = A_lQ$ . Considering that $AD$ is the internal bisector of $\angle FDE$ results that $\mathcal{A}=A^l_CC_A^l \cap A_B^lB_A^l\in AD$ . Similarly $\mathcal{B}=A^l_BB_A^l \cap C_B^lB_C^l\in BE$ and $\mathcal{C}=A^l_CC_A^l \cap C_B^lB_C^l\in CF$ . Therefore $\triangle \mathcal{ABC}$ is homothetic with $\triangle DEF$ .

nb3

Lemma 3. Let $\triangle \mathcal{ABC}$ be a triangle which is sent to $\triangle DEF$ by a homothety with center $H$ . The perpendicular bisectors $\alpha, \beta , \gamma$ of $A\mathcal{A},\ B\mathcal{B},\ C\mathcal{C}$ bound a triangle, $\triangle MNR$ , with orthocenter $O$ . This triangle is also homothetic with $\triangle ABC$ .

Proof. Clearly $\alpha \parallel BC, \beta \parallel AC$ and $\gamma \parallel AB$ . Because $AC\mathcal{AC}$ is cyclic, results $N$ is it’s center. Hence $NO \perp AC$ . Similarly $MO \perp BC$ and $RO \perp AB$ .
VTherefore $O$ is the orthocenter of $\triangle MNR$ .
nb4

Lemma 4. Let $P$ be a point and $\triangle J_1J_2J_3$ , the excentral triangle of $ABC$ . The perpendicular from $P$ to $J_1J_3,J_1J_2$ intersect $AC,AB$ in $M,N$ ,respectively. If $MN$ $\parallel J_2J_3$ then $P\in A- \rm{Nagel \ Cevian}$ .(Clearly there are analogs of the lemma when $P\in B- \rm{Nagel \ Cevian}$ and $P\in C- \rm{Nagel \ Cevian}$ )

Proof. By a homothety wrt $A$ , $P$ is sent to a point on $BC$ . And this point can only be the tangency point of $BC$ with $J_1- \rm{excircle}$ , cause the only posibility for $MN$ is to transform in the polar of $A$ wrt $J_1-\rm{excircle}$ . Therefore, $P \in A-\rm{Nagel Cevian}$ . The converse of the lemma is also true.
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Lemma 5. Let $\triangle MNR$ be a triangle with orthocenter $Na$ ,Nagel point of $\triangle ABC$ , and homothetic with the excentrel triangle, $\triangle J_1J_2J_3$ . $\triangle MNR$ is perspective with $\triangle ABC$ .

Proof. Clearly $MNa\perp J_2J_3$ , $NNa\perp J_1J_3$ , $RNa\perp J_2J_1$ . Hence, when $M$ moves on the line perpendicular from $Na$ to $J_3J_2$ , $N,R$ move accordingly on the lines perpendicular from $Na$ to $J_1J_3,J_1J_2$ , respectively. Therefore, $M\barwedge N\barwedge R$ .
V When $N$ is at the intersection of $BC$ and the perpendicular from $Na$ to $J_1J_3$ , by lemma 4, $M$ is at the intersection of $AC$ and the perpendicular from $Na$ to $J_2J_3$ . This gives that when $N$ is at the intersection of $BC$ and the perpendicular from $Na$ to $J_1J_3$ , $AM \cap AN = C$ . Also when $N$ is at infinity then $AM\cap BN=I$ . VTherefore, the intersection of $AM$ with $BN$ forms a conic passing through $A,B,C,I,Na$ . Futhermore, it is an rectangular hyperbola, because $O$ is on the isogonal locus of the conic, which is the line passing through $I$ and $X(56)$ (exsimilicenter between incenter and circumcenter). Clearly the intersection of $AM$ and $CR$ gives the same rectangular hyperbola. So $AM,BN,CR$ are concurrent on the rectangular hyperbola passing through $I,A,B,C$ .
nb5

Sondat’s Theorem. If two triangles are orthologic and perspective, then their orthology centers and homology center are on a line perpendicular to their perspective axis.

Main proof of property 1. Let $l\perp \rm{Euler \ line}$ and $\Gamma_{A} , \Gamma_{B} , \Gamma_{C}$ be the special circles studied in lemma 1, for $A^l,B^l,C^l$ . $D= \odot BOC\cap AO$ , $E= \odot AOC\cap BO$ and $F= \odot BOA\cap CO$ . Clearly $D\in \Gamma_A,E\in \Gamma_B \ \rm{and} \ F\in \Gamma_C$ .
V Let $\mathcal{ABC}$ be the triangle given by the lines $B^{l}_{A}A^{l}_{B}$ , $A^{l}_CC^{l}_{A}$ and $C^{l}_BB^{l}_{C}$ .By lemma 2 $\angle \mathcal{CAB} =180^{\circ} -2 \cdot \angle BAC= A_C^lAA_B^l$ , results $\mathcal{A} \in \Gamma_A$ . Similarly $\mathcal{B} \in \Gamma_B$ and $\mathcal{C} \in \Gamma_C$ . Also by lemma 2, $AH\cdot \mathcal{A}H = BH\cdot \mathcal{B}H= CH\cdot \mathcal{C}H$ . Hence, $H$ is the radical center of $\Gamma_{A}, \Gamma_{B}, \Gamma_{C}$ . $\clubsuit$
V $A', A"$ are the other intersections of $\odot BOC$ with $AB, \ \rm{and} \ AC$ . $B',B"$ and $C',C"$ are defined similarly. $\triangle S_1S_2S_3$ is the triangle bounded by $A'A",B'B",C'C"$
(Please see the image, for clarity). $A'A",B'B",\ \rm{and} \ C'C"$ are the perpendicular bisectors of $AD,BE, \ \rm{and} \ CF$ . So $S_1S_2S_3$ is homothetic with the $H-\rm{Pedal \ triangle}$ .
V $\triangle ABC$ is the excentral triangle of $H-\rm {pedal triangle}$ . So the excentral triangle of $\triangle S_1S_2S_3$ is homothetic with $\triangle ABC$ . Clearly $\overline{OC'B'} \perp BC, \ \overline{OA"B"} \perp AB,\ \overline{OC"A'}\perp AC$ . ( $\overline{abc}$ means that $a,b,\ \rm{and} \ c$ are collinear). Therefore, by lemma 4, $O$ is the Nagel point of $\triangle S_1S_2S_3$ . $\bigstar$
nb7

V
Let $\triangle M_1M_2M_3$ be the triangle bounded by the perpendicular bisectors of $A\mathcal{A},\ B\mathcal{B},\ C\mathcal{C}$ . By lemma 3, 5 and $\bigstar$ , results that $\triangle M_1M_2M_3$ is perspective with $\triangle S_1S_2S_3$ . Furthermore, $O_3=M_1M_2\cap S_1S_2.\ O_1=M_2M_3\cap S_2S_3,\ \rm{and}\ O_2=M_3M_1\cap S_3S_1$ , are the centers of $\Gamma_3,\ \Gamma_1,\ \rm{and} \ \Gamma_3$ , respectively. Then by Desargues’ Theorem on $\triangle M_1M_2M_3 \ \rm{and}\ \triangle S_1S_2S_3$ , results that $O_1,\ O_2,\ O_3$ are collinear.

V
By $\clubsuit$ and the conclusion of the last paragraph, the circles $\Gamma_1, \Gamma_2, \Gamma_3$ are coaxial. Let $X,Y$ be their intersections. Obviously, from lemma 1, $X-\rm{Reflection \ Triangle}$ has $l$ as a perspective axis with $\triangle ABC$ . Hence by Desargues’ Theorem the $X-\rm{Reflection\ Triangle}$ and $\triangle ABC$ are perspective. Obviously the other orthologic center between $\triangle ABC$ and $X-\rm{Reflection \ Triangle}$ is the isogonal of $X$ .

VFinally, by Sondat’s Theorem, the line containing $X$ and it’s isogonal is perpendicular to $l$ . And obviously this line passes through Euler infinity point. Hence $X\in \rm{Neuberg \ Cubic}$ , concluding the solution.( Similarly $Y\in \rm{Neuberg Cubic}$ . )