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0

1

$begingroup$

Today on the first lecture on Spectral analysis I pose a question whether we may make a sense besides this $$int f(x)textdx$$ expression also of this $$int f(x) r(textdx)$$ expression. Note that by the second I do not mean $$int f(x)textdr(x).$$ The teachre said that he cannot make sense of it, but I think that some analogy (which one is also part of my question) can be made with (infinite) but discrete summation.

real-analysis integration definite-integrals

share|cite|improve this question

asked Feb 19 at 18:21

user122424user122424

1,1582717

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $r(dx)$ and $dr(x)$ mean the same thing...
  $endgroup$
  – Surb
  Feb 19 at 18:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how would you prove or at least see this "sameness" ?
  $endgroup$
  – user122424
  Feb 19 at 18:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is nothing to prove, these are just conventional notation.
  $endgroup$
  – Surb
  Feb 19 at 18:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But instead of multiplying $r(x)$ with element dx I mean putting it into the argument of the auxilliary function $r(x)$!
  $endgroup$
  – user122424
  Feb 19 at 18:29
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have to notice that these are just conventional notation. If $r(x)=e^x$, then to see $r(dx)$ as $e^dx$ doesn't make sense. The notation $dr(x)$ is common when $r$ is derivable, and if not, the notation $r(dx)$ is commonly used... but there are no rules ! In the case of my example, $r(dx)$ and $dr(x)$ both refer to the measure $e^xdx$.
  $endgroup$
  – Surb
  Feb 19 at 18:32
  
  

  
  
  
  

 | 
show 4 more comments

0

1

$begingroup$

Today on the first lecture on Spectral analysis I pose a question whether we may make a sense besides this $$int f(x)textdx$$ expression also of this $$int f(x) r(textdx)$$ expression. Note that by the second I do not mean $$int f(x)textdr(x).$$ The teachre said that he cannot make sense of it, but I think that some analogy (which one is also part of my question) can be made with (infinite) but discrete summation.

real-analysis integration definite-integrals

share|cite|improve this question

asked Feb 19 at 18:21

user122424user122424

1,1582717

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $r(dx)$ and $dr(x)$ mean the same thing...
  $endgroup$
  – Surb
  Feb 19 at 18:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how would you prove or at least see this "sameness" ?
  $endgroup$
  – user122424
  Feb 19 at 18:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is nothing to prove, these are just conventional notation.
  $endgroup$
  – Surb
  Feb 19 at 18:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But instead of multiplying $r(x)$ with element dx I mean putting it into the argument of the auxilliary function $r(x)$!
  $endgroup$
  – user122424
  Feb 19 at 18:29
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have to notice that these are just conventional notation. If $r(x)=e^x$, then to see $r(dx)$ as $e^dx$ doesn't make sense. The notation $dr(x)$ is common when $r$ is derivable, and if not, the notation $r(dx)$ is commonly used... but there are no rules ! In the case of my example, $r(dx)$ and $dr(x)$ both refer to the measure $e^xdx$.
  $endgroup$
  – Surb
  Feb 19 at 18:32
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Today on the first lecture on Spectral analysis I pose a question whether we may make a sense besides this $$int f(x)textdx$$ expression also of this $$int f(x) r(textdx)$$ expression. Note that by the second I do not mean $$int f(x)textdr(x).$$ The teachre said that he cannot make sense of it, but I think that some analogy (which one is also part of my question) can be made with (infinite) but discrete summation.

real-analysis integration definite-integrals

share|cite|improve this question

asked Feb 19 at 18:21

user122424user122424

1,1582717

$endgroup$

share|cite|improve this question

asked Feb 19 at 18:21

user122424user122424

1,1582717

share|cite|improve this question

asked Feb 19 at 18:21

user122424user122424

1,1582717

asked Feb 19 at 18:21

user122424user122424

1,1582717

asked Feb 19 at 18:21

user122424user122424

1,1582717

2 Answers
2

active

oldest

votes

2

$begingroup$

As many have commented, $int f(x)mu(dx)$ is yet another variant notation for what is more often written as $int f dmu$ or $int f(x) dmu$ or $int f(x) dmu(x)$. Perhaps probabilists use it more often than other people. Doob's 1952 Stochastic Processes uses it. Theorem 9.1 in Chap. 1 says "... making the obvious notational conventions
$$mathbf Ey=int_Omega y(omega')mathbf P(domega', omega)."$$
Here his $mathbf P$ is a family of measures, parameterized by the second argument, $omega$, so for a set $A$ and value of $omega$, $mathbf P(A,omega)$ is a number.

This convention is most useful when one has multiple integrals and more than one possible variable of integration.

As many comments indicate, there is a perverse kind of logic in thinking of $int f(x)mu(dx)$ as suggesting a kind of Riemann sum, where one adds up rectangles of
height $f(x)$ and base $[x,x+dx)$, weighting each with measure $mu([x,x+dx))$, finally taking $dx$ as shorthand for the short interval $[x,x+dx)$.

But of course it is just another notational convention, an idiomatic expression in the language of probabalists & other analysts.

Added 3 March 2019: Dunford and Schwartz use this notation in Linear Operators, Part 1, II.2.13, p. 108.

share|cite|improve this answer

edited Mar 10 at 18:25

answered Feb 19 at 19:56

kimchi loverkimchi lover

11.1k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about this interpretation: $$int_a^b f(x)r(textdx)=lim_textdxto 0sum_x=a^b f(x)cdot r(textdx)$$
  $endgroup$
  – user122424
  Feb 19 at 20:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't make sense ! Either $r$ is a bounded variation function and $int_a^b f(x)dx=lim_nto infty sum_k=1^nf(x_k)(mu(t_k+1)-mu(t_k))$ for $t_0,...,t_n$ a partition of $[a,b]$ s.t. $max_i|t_i+1-t_i|to 0$ as $nto infty $, or $r$ is a counting measure, and thus $int_a^b f(x)r(dx)=sum_kin mathcal Df(k)r(k)$ for $mathcal D=kin [a,b]mid r(k)>0$ countable. As kimchi lover well explained in his answer, you really have to think $r(dx)$ as $r([x,x+dx])$ for $dx$ very small.
  $endgroup$
  – Surb
  Feb 19 at 20:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Notice that if $F$ is increasing, then $F(dx)=F([x,x+dx])=[F(x),F(x+dx)]$ which have length $F(x+dx)-F(x)$ what we commonly write as $dF(x)$... maybe this is a motivation for using $dF(x)$ and $F(dx)$ for the same thing. @user122424
  $endgroup$
  – Surb
  Feb 19 at 20:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I really cannot imagine which part of my formula above with $lim$ makes no sense??
  $endgroup$
  – user122424
  Feb 20 at 12:02
  

  
  
  
  

add a comment | 

0

$begingroup$

If $f(0) = 0$ and $f'(0)$ is defined then we can give $f(mathrmdx)$ a quite natural meaning:
$f(mathrmdx) := f'(0) , mathrmdx.$

Generally, if $f'(x_0)$ is defined then we take $f(x_0 + mathrmdx) := f(x_0) + f'(x_0) , mathrmdx.$ This is motivated by the formal identity
$$fracf(x_0+mathrmdx) - f(x_0)mathrmdx = fracmathrmdfmathrmdx(x_0) = f'(x_0).$$

share|cite|improve this answer

answered Feb 19 at 22:40

md2perpemd2perpe

8,19611028

$endgroup$

add a comment | 

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2

$begingroup$

As many have commented, $int f(x)mu(dx)$ is yet another variant notation for what is more often written as $int f dmu$ or $int f(x) dmu$ or $int f(x) dmu(x)$. Perhaps probabilists use it more often than other people. Doob's 1952 Stochastic Processes uses it. Theorem 9.1 in Chap. 1 says "... making the obvious notational conventions
$$mathbf Ey=int_Omega y(omega')mathbf P(domega', omega)."$$
Here his $mathbf P$ is a family of measures, parameterized by the second argument, $omega$, so for a set $A$ and value of $omega$, $mathbf P(A,omega)$ is a number.

This convention is most useful when one has multiple integrals and more than one possible variable of integration.

As many comments indicate, there is a perverse kind of logic in thinking of $int f(x)mu(dx)$ as suggesting a kind of Riemann sum, where one adds up rectangles of
height $f(x)$ and base $[x,x+dx)$, weighting each with measure $mu([x,x+dx))$, finally taking $dx$ as shorthand for the short interval $[x,x+dx)$.

But of course it is just another notational convention, an idiomatic expression in the language of probabalists & other analysts.

Added 3 March 2019: Dunford and Schwartz use this notation in Linear Operators, Part 1, II.2.13, p. 108.

share|cite|improve this answer

edited Mar 10 at 18:25

answered Feb 19 at 19:56

kimchi loverkimchi lover

11.1k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about this interpretation: $$int_a^b f(x)r(textdx)=lim_textdxto 0sum_x=a^b f(x)cdot r(textdx)$$
  $endgroup$
  – user122424
  Feb 19 at 20:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't make sense ! Either $r$ is a bounded variation function and $int_a^b f(x)dx=lim_nto infty sum_k=1^nf(x_k)(mu(t_k+1)-mu(t_k))$ for $t_0,...,t_n$ a partition of $[a,b]$ s.t. $max_i|t_i+1-t_i|to 0$ as $nto infty $, or $r$ is a counting measure, and thus $int_a^b f(x)r(dx)=sum_kin mathcal Df(k)r(k)$ for $mathcal D=kin [a,b]mid r(k)>0$ countable. As kimchi lover well explained in his answer, you really have to think $r(dx)$ as $r([x,x+dx])$ for $dx$ very small.
  $endgroup$
  – Surb
  Feb 19 at 20:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Notice that if $F$ is increasing, then $F(dx)=F([x,x+dx])=[F(x),F(x+dx)]$ which have length $F(x+dx)-F(x)$ what we commonly write as $dF(x)$... maybe this is a motivation for using $dF(x)$ and $F(dx)$ for the same thing. @user122424
  $endgroup$
  – Surb
  Feb 19 at 20:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I really cannot imagine which part of my formula above with $lim$ makes no sense??
  $endgroup$
  – user122424
  Feb 20 at 12:02
  

  
  
  
  

add a comment | 

2

$begingroup$

As many have commented, $int f(x)mu(dx)$ is yet another variant notation for what is more often written as $int f dmu$ or $int f(x) dmu$ or $int f(x) dmu(x)$. Perhaps probabilists use it more often than other people. Doob's 1952 Stochastic Processes uses it. Theorem 9.1 in Chap. 1 says "... making the obvious notational conventions
$$mathbf Ey=int_Omega y(omega')mathbf P(domega', omega)."$$
Here his $mathbf P$ is a family of measures, parameterized by the second argument, $omega$, so for a set $A$ and value of $omega$, $mathbf P(A,omega)$ is a number.

This convention is most useful when one has multiple integrals and more than one possible variable of integration.

As many comments indicate, there is a perverse kind of logic in thinking of $int f(x)mu(dx)$ as suggesting a kind of Riemann sum, where one adds up rectangles of
height $f(x)$ and base $[x,x+dx)$, weighting each with measure $mu([x,x+dx))$, finally taking $dx$ as shorthand for the short interval $[x,x+dx)$.

But of course it is just another notational convention, an idiomatic expression in the language of probabalists & other analysts.

Added 3 March 2019: Dunford and Schwartz use this notation in Linear Operators, Part 1, II.2.13, p. 108.

share|cite|improve this answer

edited Mar 10 at 18:25

answered Feb 19 at 19:56

kimchi loverkimchi lover

11.1k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about this interpretation: $$int_a^b f(x)r(textdx)=lim_textdxto 0sum_x=a^b f(x)cdot r(textdx)$$
  $endgroup$
  – user122424
  Feb 19 at 20:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't make sense ! Either $r$ is a bounded variation function and $int_a^b f(x)dx=lim_nto infty sum_k=1^nf(x_k)(mu(t_k+1)-mu(t_k))$ for $t_0,...,t_n$ a partition of $[a,b]$ s.t. $max_i|t_i+1-t_i|to 0$ as $nto infty $, or $r$ is a counting measure, and thus $int_a^b f(x)r(dx)=sum_kin mathcal Df(k)r(k)$ for $mathcal D=kin [a,b]mid r(k)>0$ countable. As kimchi lover well explained in his answer, you really have to think $r(dx)$ as $r([x,x+dx])$ for $dx$ very small.
  $endgroup$
  – Surb
  Feb 19 at 20:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Notice that if $F$ is increasing, then $F(dx)=F([x,x+dx])=[F(x),F(x+dx)]$ which have length $F(x+dx)-F(x)$ what we commonly write as $dF(x)$... maybe this is a motivation for using $dF(x)$ and $F(dx)$ for the same thing. @user122424
  $endgroup$
  – Surb
  Feb 19 at 20:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I really cannot imagine which part of my formula above with $lim$ makes no sense??
  $endgroup$
  – user122424
  Feb 20 at 12:02
  

  
  
  
  

add a comment | 

$begingroup$

As many have commented, $int f(x)mu(dx)$ is yet another variant notation for what is more often written as $int f dmu$ or $int f(x) dmu$ or $int f(x) dmu(x)$. Perhaps probabilists use it more often than other people. Doob's 1952 Stochastic Processes uses it. Theorem 9.1 in Chap. 1 says "... making the obvious notational conventions
$$mathbf Ey=int_Omega y(omega')mathbf P(domega', omega)."$$
Here his $mathbf P$ is a family of measures, parameterized by the second argument, $omega$, so for a set $A$ and value of $omega$, $mathbf P(A,omega)$ is a number.

This convention is most useful when one has multiple integrals and more than one possible variable of integration.

As many comments indicate, there is a perverse kind of logic in thinking of $int f(x)mu(dx)$ as suggesting a kind of Riemann sum, where one adds up rectangles of
height $f(x)$ and base $[x,x+dx)$, weighting each with measure $mu([x,x+dx))$, finally taking $dx$ as shorthand for the short interval $[x,x+dx)$.

But of course it is just another notational convention, an idiomatic expression in the language of probabalists & other analysts.

Added 3 March 2019: Dunford and Schwartz use this notation in Linear Operators, Part 1, II.2.13, p. 108.

share|cite|improve this answer

edited Mar 10 at 18:25

answered Feb 19 at 19:56

kimchi loverkimchi lover

11.1k31229

$endgroup$

share|cite|improve this answer

edited Mar 10 at 18:25

answered Feb 19 at 19:56

kimchi loverkimchi lover

11.1k31229

answered Feb 19 at 19:56

kimchi loverkimchi lover

11.1k31229

answered Feb 19 at 19:56

kimchi loverkimchi lover

11.1k31229

0

$begingroup$

If $f(0) = 0$ and $f'(0)$ is defined then we can give $f(mathrmdx)$ a quite natural meaning:
$f(mathrmdx) := f'(0) , mathrmdx.$

Generally, if $f'(x_0)$ is defined then we take $f(x_0 + mathrmdx) := f(x_0) + f'(x_0) , mathrmdx.$ This is motivated by the formal identity
$$fracf(x_0+mathrmdx) - f(x_0)mathrmdx = fracmathrmdfmathrmdx(x_0) = f'(x_0).$$

share|cite|improve this answer

answered Feb 19 at 22:40

md2perpemd2perpe

8,19611028

$endgroup$

add a comment | 

0

$begingroup$

If $f(0) = 0$ and $f'(0)$ is defined then we can give $f(mathrmdx)$ a quite natural meaning:
$f(mathrmdx) := f'(0) , mathrmdx.$

Generally, if $f'(x_0)$ is defined then we take $f(x_0 + mathrmdx) := f(x_0) + f'(x_0) , mathrmdx.$ This is motivated by the formal identity
$$fracf(x_0+mathrmdx) - f(x_0)mathrmdx = fracmathrmdfmathrmdx(x_0) = f'(x_0).$$

share|cite|improve this answer

answered Feb 19 at 22:40

md2perpemd2perpe

8,19611028

$endgroup$

add a comment | 

$begingroup$

If $f(0) = 0$ and $f'(0)$ is defined then we can give $f(mathrmdx)$ a quite natural meaning:
$f(mathrmdx) := f'(0) , mathrmdx.$

Generally, if $f'(x_0)$ is defined then we take $f(x_0 + mathrmdx) := f(x_0) + f'(x_0) , mathrmdx.$ This is motivated by the formal identity
$$fracf(x_0+mathrmdx) - f(x_0)mathrmdx = fracmathrmdfmathrmdx(x_0) = f'(x_0).$$

share|cite|improve this answer

answered Feb 19 at 22:40

md2perpemd2perpe

8,19611028

$endgroup$

share|cite|improve this answer

answered Feb 19 at 22:40

md2perpemd2perpe

8,19611028

answered Feb 19 at 22:40

md2perpemd2perpe

8,19611028

answered Feb 19 at 22:40

md2perpemd2perpe

8,19611028