This statement "the probability that a continuous variable exactly equals to a given value is zero" is commonly used.  is this an axiom, a property of continuous variable, a rule of integration or a conclusion of measure theory?  please provide a citation of a textbook or formally publication.  the  itself some answer linked is being tagged "unclear sources"
6674
My understanding is that a random variable is actually a function $X: \Omega \to T$, where $\Omega$ is the sample space of some random experiment and $T$ is the set from which the possible values of the random variable are taken.  Regarding the set of values that the random variable can actually take, it is the image of the function $X$. If the image is finite, then $X$ must be a discrete random variable. However, if it is an infinite set, then $X$ may or may not be a continuous random variable. Whether it is depends on whether the image is countable or not. If it is countable, then $X$ is a discrete random variable; whereas if it is not, then $X$ is  continuous.  Assuming that my understanding is correct, why does the fact that the image is uncountable imply that $Pr(X = x) = 0$.  I would have thought that the fact that the image is infinite, regardless of whether it is countable or not, would already imply that $Pr(X = x) = 0$ since if it is infinite, then the domain $\Omega$ must also be infinite, and therefore  $$Pr(X = x) = \frac{\text{# favorable outcomes}}{\text{# possible outcomes}} = \frac{\text{# outcomes of the experiment where X = x}}{|\Omega|} = \frac{\text{# outcomes of the experiment where X = x}}{\infty} = 0$$  What is wrong with my argument?  Why does the probability that a continuous random variable takes on a specific value actually equal zero?
22587
Let $\Omega$ be a measure space with measure $\mu$. (Here, a measure is only meant to be countable additive!)  Consider a subset $A\in\Sigma$.  Then according to the  it is an atom if: $$(1)\quad\forall E\subseteq A:\quad\mu(E)&lt;\mu(A)\implies\mu(E)=0\quad(\mu(A)&gt;0)$$ and according to the  it is an atom if: $$(2)\quad\forall E\in\Sigma:\quad\mu(E\cap A)=0\lor\mu(E^c\cap A)=0\quad(\mu(A)&gt;0)$$  Now, these definitions agree for the really trivial cases: $$\mu\equiv0$$ $$\mu\equiv\infty$$ but they differ for the less trivial case: $$\quad\mu(E\neq\varnothing):=\infty,\,\mu(\varnothing):=0$$ namely the atoms are w.r.t. (1) all nonempty subsets whereas w.r.t. (2) only the singletons.  Excluding this pathological case, are the definitions equivalent? (More precisely, assume there exists a measurable subset $0&lt;\mu(F)&lt;\infty$.)  So far I checked that: $$(2)\implies(1):\quad\mu(E)&lt;\mu(A)\implies\mu(A\setminus E)&gt;0\implies\mu(E)=0$$ but what about the other direction?
68189
Prove that if $f\colon\mathbb{R}\to\mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ for all $x,y$, and $f$ is continuous at $0$, then it is continuous everywhere.   If there exists $c \in \mathbb{R}$ such that $f(c) = 0$, then $$f(x + c) = f(x)f(c) = 0.$$ As every real number $y$ can be written as $y = x + c$ for some real $x$, this function is either everywhere zero or nowhere zero. The latter case is the interesting one. So let's consider the case that $f$ is not the constant function $f = 0$.  To prove continuity in this case, note that for any $x \in \mathbb{R}$ $$f(x) = f(x + 0) = f(x)f(0) \implies f(0) = 1.$$  Continuity at $0$ tells us that given any $\varepsilon_0 &gt; 0$, we can find $\delta_0 &gt; 0$ such that $|x| &lt; \delta_0$ implies $$|f(x) - 1| &lt; \varepsilon_0.$$  Okay, so let $c \in \mathbb{R}$ be fixed arbitrarily (recall that $f(c)$ is nonzero). Let $\varepsilon &gt; 0$. By continuity of $f$ at $0$, we can choose $\delta &gt; 0$ such that $$|x - c| &lt; \delta\implies |f(x - c) - 1| &lt; \frac{\varepsilon}{|f(c)|}.$$  Now notice that for all $x$ such that $|x - c| &lt; \delta$, we have $$\begin{align*} |f(x) - f(c)| &amp;= |f(x - c + c) - f(c)|\\  &amp;= |f(x - c)f(c) - f(c)|\\  &amp;= |f(c)| |f(x - c) - 1|\\  &amp;\lt |f(c)| \frac{\varepsilon}{|f(c)|}\\   &amp;= \varepsilon. \end{align*}$$ Hence $f$ is continuous at $c$. Since $c$ was arbitrary, $f$ is continuous on all of $\mathbb{R}$.  Is my procedure correct?
21944
Pretty much the problem asks if a function is continuous at the point $c$ and $f(c) &gt; 0$ then there exists a $d &gt; 0$ such that $\forall x$, $f(x) &gt; 0$ with $|x-c| &lt; d$.   I can understand what the problem means. That if a function is positive at a point then there is another point that's really close that'll also be positive. I can not prove this in a formal way though. I've tried using the intermediate value theorem and I do not know hot to implement it. Any ideas?
33427
Using the definition of continuity and the $\epsilon$-$\delta$ definition of a limit, how would you proceed to show that the following statement holds true?  Let $f : \Bbb{R} \rightarrow \Bbb{R}$ be a continuos function at $x=0$, with the property that if $x\neq 0$ then $f(x)\geq 0$. Then $f(0) \geq 0$.  My attempt so far has been to assume that $f(0) \lt 0$, which means that $$f(0)=\lim_{x\to 0} f(x)\lt 0,$$ because of the continuity at the given point. Furthermore, I think this means that when $x$ approaches $0$ from both sides, at some point the value $f(x)\lt 0$, thus forming a contradiction.   I can't seem to see what to do afterwards. Is this a way of approaching the problem? And how will the $\epsilon-\delta$ definition come into play if my approach is usable?
88229
Consider a probability space $(\Omega, \mathcal{F}, P)$ where $\Omega$ is the sample space, $\mathcal{F}$ is the $\sigma$-algebra of $\Omega$, and $P$ is the probability measure. Let $X:\Omega\to\mathbb{R}$ be a random variable, inducing a $\sigma$-algebra $\mathcal{F}_X=\{X^{-1}(B)\mid B\in\mathcal{B}\}$ where $\mathcal{B}$ is the Borel algebra on $\mathbb{R}$.   Suppose we have another random variable $Y$ as well as the induced $\sigma$-algebra $\mathcal{F}_Y$, the conditional expectation can be defined as $E[Y|X]\stackrel{\text{a.s.}}{=}g$ where $g$ is determined by $E[YI_A]=\int_A g(s)P(s)\,\mathrm{d}s, \forall A\in \mathcal{F}_X$ according to Radon–Nikodym theorem.  I wonder what will happen if $\mathcal{F}_X=\mathcal{F}$ already? It looks like $E[Y|X]=Y$ so $X$ is kind of independent of any arbitrary random variable but I am not sure. Or, can anybody give an example of this kind of $X$? Thank you.  Some references:  by Mark J. Schervish.
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The question is the following: Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0$, $f'(x)$ exists. If $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists?  My intuition is that this does not have to hold in general. However, I keep finding counter examples such that $f'(0)$ exists where $\lim_{x \rightarrow 0} f'(x) = L$ does not exist, which is obviously not the original question.
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I wanna prove that   "if $f: \mathbb{R}^n \to \mathbb{R}$ is continuous and satisfies $f=0$ almost everywhere (in the sense of Lebesgue measure), then, $f=0$ everywhere."  I am confident that the statement is true, but stuck with the proof. Also, is the statement true if the domain $\mathbb{R}^n$ is restricted to $\Omega \subseteq \mathbb{R}^n$ that contains a neighborhood of the origin "$0$"?
121480
I just had an (intellectual) panic attack.     A continuous random variable that follows a uniform in a closed interval $U(a,b)$: a comfortably familiar statistical concept.   A continuous uniform r.v. having support over the extended reals (half or whole): not an r.v. proper, but a basic Bayesian concept for an improper prior, useful and applicable. A discrete uniform taking a finite number of values: let's throw a geodesic dome, no big deal.     But what about a function that has as its domain all the rationals that are included in a closed interval with integer bounds (start with the $[0,1]$ if you wish)? And we want to use it in a probabilistic framework, requiring that each possible value has equal probability with all the others?   The number of possible values are countably infinite (which characterizes a lot of discrete distributions), but how to express the probability of a single value given that we want probabilities equal?  Can we say-show-prove that such an entity is (is not) a random variable?    If not, is this another incarnation (perhaps already well-known) of an "improper prior"?  Is it possible that this entity is in some well-defined sense, however special,  "equivalent" to a continuous uniform r.v.? Or I just committed a cardinal(ity) sin?  It appears that the fact that the domain is a closed interval does not let me let go. Bounded things are usually manageable.  The questions are many in order to be indicative of the internal maelstrom- I am not asking to get answers to each one of them.    At any time that I may come up with any insights, I will update.    UPDATE: the present question just acquired a constructivist sequel
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This theorem seems to hold great potential for an intuitive explanation, yet I'm still struggling for it.  Basically, it is saying that if $y = F(x)$, then y would have distribution $Unif[0,1]$. Intuitively, why?  Many thanks!
61825
Suppose $f$ is continuous on $[0,1]$ and   $$ \int\limits_0^1 f(x) x^n d x = 0 $$   for all $n=0,1,2,...$. Then, $f(x) = 0$  I was trying to use by parts  $$ 0 = f(x) \frac{x^{n+1}}{n+1} \bigg|_0^1 - \int\limits_0^1 \frac{x^{n+1}}{n+1} f(x) dx $$  Thus,  $$ \frac{f(1)}{n+1} = \int\limits_0^1 \frac{ x^{n+1} f(x) dx }{n+1} \implies f(1) = \int\limits_0^1 x^{n+1} f(x) dx $$  but here I am stuck. Maybe my idea is not the best one? How would you approach this problem?
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Can an observed event in fact be of zero probability?  Of course, I know that there exist non-empty events of zero probability. Mu question is the reverse: given that we have observed an event (and we have no other information about it, just the fact that it has been observed), is it possible that the event has in fact zero probability? Or does an observation necessarily mean that the probability is strictly positive?    Example context for the question:  Suppose $x_i$ (countably many) are i.i.d on $[0,1]$, but we do not know the distribution they come from. It may be uniform on $[0,1]$, may be discrete, may be any legitimate distribution (discrete or continuous). Just imagine we have some sort of a machine that shows us one by one a list of randomly drawn numbers between $0$ and $1$. We are comparing the observed numbers one by one to some special number chosen beforehand, for example $\frac12$. Now, given that at some iteration we have observed that special number at least once, does that mean that $\frac12$ has some positive probability under that (unknown) distribution? And if the probability can be zero, can we nevertheless say that we will necessarily observe $\frac12$ again later, if we continue the experiment ad infimum?  Also, disregard the "real world limitations" such as an inability to produce truly uniformly distibuted numbers, or rounding errors or any such thing.
120426
I am really satisfied that $(x \to a) ≠(x=a)$ and if that is not right , Then all the process of $Limits$ is dividing by zero and that is a crime.  Since $(x \to a) + h = (x=a)$ , $h ≠ 0$,So Why does $f(x \to a) = f(x = a)$ ?  NOTE:I am talking about continuous function.
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If $f(x)$ is a density function and $F(x)$ is a distribution function of a random variable $X$ then I understand that the expectation of x is often written as:   $$E(X) = \int x f(x) dx$$  where the bounds of integration are implicitly $-\infty$ and $\infty$. The idea of multiplying x by the probability of x and summing makes sense in the discrete case, and it's easy to see how it generalises to the continuous case.  However, in Larry Wasserman's book All of Statistics he writes the expectation as follows:  $$E(X) = \int x dF(x)$$  I guess my calculus is a bit rusty, in that I'm not that familiar with the idea of integrating over functions of $x$ rather than just $x$.   What does it mean to integrate over the distribution function? Is there an analogous process to repeated summing in the discrete case? Is there a visual analogy?   UPDATE: I just found the following extract from Wasserman's book (p.47):     The notation $\int x d F(x)$ deserves some comment. We use it merely   as a convenient unifying notation so that we don't have to write   $\sum_x x f(x)$ for discrete random variables and $\int x f(x) dx$ for   continuous random variables, but you should be aware that $\int x d F(x)$  has a precise meaning that is discussed in a real analysis   course.   Thus, I would be interested in any insights that could be shared about what is the precise meaning that would be discussed in a real analysis course?
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A function is "differentiable" if it has a derivative. A function is "continuous" if it has no sudden jumps in it.  Until today, I thought these were merely two equivalent definitions of the same concept. But I've read some stuff today which seems to be claiming that this is not the case.  The obvious next question is "why?" Apparently somebody has already asked:    Several answers were given, but I don't understand any of them. In particular, Wikipedia and one of the replies above both claim that $|x|$ has no derivative. Can anyone explain this extremely unexpected result?  Edit: Apparently some people dislike the fact that this is non-obvious to me. To be clear: I am not saying that the result is untrue. (I'm sure many great mathematicians have analysed the question very carefuly and are quite sure of the answer.) I am saying that it is extremely perplexing. (As a general rule, mathematics has a habit of doing that. Which is one of the reasons why we demand proof of everything.)  In particular, can anyone explain precisely why the derivative of $|x|$ at zero is not simply zero? After all, the function is neither increasing nor decreasing, which ought to mean the derivative is zero. Alternatively, the expression  $$\frac{|x + a| - |x - a|}{a}$$  becomes closer and closer to zero as $a$ becomes closer to zero when $x=0$. (In fact, it is exactly zero for all $a$!) Is that not how derivatives work?  Several answers have suggested that the derivative is not defined here "because there would be a jump in the derivative at that point". This seems to assert that a continuous function must never have a discontinuous derivative; I'm not convinced that this is the case. Can anyone confirm or refuse this argument?
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Let $g:[0,1] \rightarrow \mathbb R$ be a continuous function and assume that    $$ \int_{0}^{1} g(x) \phi'(x) dx = 0 $$   for all continuously differentiable functions $\phi: [0,1] \rightarrow \mathbb R$, where $\phi (0) = \phi(1) = 0$. Show that $g(x) = c$ for all $x \in [0,1]$ for some constant $c \in \mathbb R$.    My thoughts are applying integration by parts with $G(t) = \int_{0}^{t} g(x) dx $, since now we get  $$ 0 = \int_{0}^{1} G(t) \phi'(t) dt  + \int_{0}^{1} g(t) \phi (t) dt $$  which doesn't really go far. The second thought was applying MVT for definite integrals, which doesn't go far either as the conditions are hardly satisfied.   Any hints?
113366
I am having trouble with the following problem:      $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$:   $$\int_{[0,a]}f\,dm=0.$$   Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).    I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.  My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.  Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.  Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions?