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# Is Such a Function Measurable in the Product Measure Space? 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. â€” â€” â€” â€” â€” â€”

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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