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2021-11-21

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You are missing some hypothesis on $x(t,omega)$. Assuming that $(t,omega) to x(t,omega)$ is measurable you can prove this as follows: let $g_n(t,omega)=g(t,frac i 2^n)$ if $x(t,omega) in [frac i-1 2^n,frac i 2^n)$. Then each $g_n$ is measurable and $g_nto g$ pointwise.

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**The semifinite portion of a measure $mu$**

I agree with you that $mu_1$ as defined in Royden-Fitzpatrick does not yield a measure. An example that shows this is obtained by considering the measure on $mathbbN$ defined on all subsets $A subseteq mathbb N$ by $$ mu(A) = begincases 0, & textif A = emptyset, 1, & textif A = 0 infty, & textotherwise. endcases $$ For this measure $mu$ we have $mu_1(emptyset) = 0$, $mu_1(0) = 1$ and $mu_1(A) = 0$ for all other $A subseteq mathbbN$, so $mu_1$ is obviously not a measure.The usual definition of the semi-finite version $mu_

m sf$ of a measure $mu$ on $(X,Sigma)$ is given by $$ mu_

m sf(E) = supmu(A) mid A subseteq E text measurable, mu(A) lt infty quad textfor E in Sigma $$ In this definition it is clear that $mu_

m sf$ is non-negative, that $mu_

m sf(emptyset) = 0$ and that $mu(E) =mu_

m sf(E)$ whenever $E in Sigma$ has finite $mu$-measure. Using this, it is not hard to check that $mu_

m sf$ is $sigma$-additive on $Sigma$, hence it is a measure on $(X,Sigma)$. Moreover, $mu_

m sf$ is semi-finite and we have $mu_

m sf = mu$ if and only if $mu$ is semi-finite.One can also show that $mu_

m sf$ and $mu$ have the same integrable functions (up to $mu_

m sf$-null sets) and that $mu_

m sf$ is complete only if $mu$ is complete.

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**bounding the measure of a disjoint union under a product measure**

What your assumption is saying is that the section$$ E^(x)=y mid (x,y)in E $$ is given by $E^(x)=E_x$. Now, your prblem states that $P=P_1times P_2$ is some product measure, so that we can use the Fubini-Tonelli theorem on the indicator function $1_E$ to conclude$$ P(E)=int 1_E (x,y) dP(x,y) =int int 1_E(x,y)dP_2(y) dP_1(x)=int P_2 (E_x) dP_1(x) geq int 1/2 dP_1(x)=1/2. $$

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**Integral of $1$ minus Dirichlet's function**

The rationals are a set of Lebesgue measure zero, so they do not impact the value of integrals.Here's how you might see that. Start with an epsilon > 0. You can count the rationals, so put an interval of length epsilon/2 over the first rational. Put an interval of length epsilon/4 over the second rational, etc. You've created a sequence of intervals of total length epsilon that cover the rationals, so they must have measure less than epsilon. But epsilon was arbitrary, so they must have measure zero

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**Measure theoretic formulation**

A probability mass function of a discrete random variable X displaystyle X can be seen as a special case of two more general measure theoretic constructions: the distribution of X displaystyle X and the probability density function of X displaystyle X with respect to the counting measure. We make this more precise below. Suppose that ( A , A , P ) displaystyle (A,mathcal A,P) is a probability space and that ( B , B ) displaystyle (B,mathcal B) is a measurable space whose underlying -algebra is discrete, so in particular contains singleton sets of B displaystyle B . In this setting, a random variable X : A B displaystyle Xcolon Ato B is discrete, provided that its image is countable. The pushforward measure X ( P ) displaystyle X_*(P) -called a distribution of X displaystyle X in this context-is a probability measure on B displaystyle B whose restriction to singleton sets induces a probability mass function f X : B R displaystyle f_Xcolon Bto mathbb R , since f X ( b ) = P ( X 1 ( b ) ) = [ X ( P ) ] ( b ) displaystyle f_X(b)=P(X^-1(b))=[X_*(P)](b) for each b B displaystyle bin B . Now, suppose that ( B , B , ) displaystyle (B,mathcal B,mu ) is a measure space equipped with the counting measure . The probability density function f displaystyle f of X displaystyle X with respect to the counting measure, if it exists, is the Radon-Nikodym derivative of the pushforward measure of X displaystyle X (with respect to the counting measure), so f = d X P / d displaystyle f=dX_*P/dmu and f displaystyle f is a function from B displaystyle B to the non-negative reals. Hence for any b B displaystyle bin B , we have that P ( X = b ) = P ( X 1 ( b ) ) := X 1 ( b ) d P = displaystyle P(X=b)=P(X^-1(b)):=int _X^-1(b)dP= b f d = f ( b ) , displaystyle int _bfdmu =f(b), demonstrating that f displaystyle f is in fact a probability mass function. When there is a natural order among the potential outcomes x displaystyle x , it may be convenient to assign numerical values to them (or n-tuples in case of a discrete multivariate random variable), and to consider also values not in the image of X displaystyle X . That is, f X displaystyle f_X may be defined for all real numbers and f X ( x ) = 0 displaystyle f_X(x)=0 for all x X ( S ) displaystyle x

otin X(S) as shown in the figure. The image of X displaystyle X has a countable subset on which the probability mass function f X ( x ) displaystyle f_X(x) is one. Consequently, the probability mass function is zero for all but a countable number of values of x displaystyle x . The discontinuity of probability mass functions is related to the fact that the cumulative distribution function of a discrete random variable is also discontinuous. If X displaystyle X is a discrete random variable, then P ( X = x ) = 1 displaystyle P(X=x)=1 means that the casual event ( X = x ) displaystyle (X=x) is certain (it is true in the 100% of the occurrences); on the contrary, P ( X = x ) = 0 displaystyle P(X=x)=0 means that the casual event ( X = x ) displaystyle (X=x) is always impossible. This statement is not true for a continuous random variable X displaystyle X , for which P ( X = x ) = 0 displaystyle P(X=x)=0 for any possible x displaystyle x : in fact, by definition, a continuous random variable can have an infinite set of possible values, and thus the probability that it has a single particular value x is equal to 1 = 0 displaystyle tfrac 1infty =0 . Discretization is the process of converting a continuous random variable into a discrete one.

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