Talk:Dirac measure

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 Field: Analysis

I suggest that the element x in the definition of the Dirac measure be replaced by a different element (say, a), so that it is easy to distinguish it from the set X. Currently, the use of both the uppercase and the lowercase symbols is confusing, especially when one is used in subscripts.

I'll wait 3-4 days to see if there are any comments before making this change.

Arcenciel 05:20, 2 November 2005 (UTC)

is the Dirac measure really an atomic measure?[edit]

Consider a delta sequence Dn(x)and consider an integrable function f(.) such that f(0+) and f(0-) exist. Then the integrals of Dn(x)f(x)dx converge to [f(0+)+f(0-)]/2, not to f(0). In fact the value of f(0) is of no influence on the limit of these integrals.

Leocat 30 September 2006

This is not a contradiction. The delta sequence D_n converges weakly (or vaguely). This means that, for every bounded continuous function (or cont. function with compact support) f, the sequence of integrals converges to the integral w.r.t. the limit-measure:
\int f(x)D_n(x)\, dx \to \int f(x) \, d \delta_0(x) = f(0)
But this need not be true for non-continuous functions. (More exactly, it is true for all functions which are continuous everywhere except on a set which is a null set for the limit measure. This is not the case for the function you mentioned.)--Trigamma 12:16, 15 January 2007 (UTC)

Restrictions on sigma-algebra[edit]

The current article says that any sigma-algebra does. However, since the singleton {x} is to be assigned a measure, ít clearly must be included in the sigma-algebra, right?

--Kaba3 (talk) 18:50, 27 August 2010 (UTC)

Aah.. I see, there is an error in the article when it claims that singleton {x} is assigned measure 1. Indeed, that singleton does not even need to be measurable. I'll fix this.

--Kaba3 (talk) 18:55, 27 August 2010 (UTC)