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The image of the Lebesgue measure on $[0, t]$ under the map $w$ (the pushforward measure) has a density $L_t(·)$. Thus,

$ \int_0^t f(w(s)) \, \mathrm{d}s = \int_{-\infty}^{+\infty} f(x) L_t(x) \, \mathrm{d}x$

for a wide class of functions $f$ (namely: all continuous functions; all locally integrable functions; all non-negative measurable functions). The density $L_t$ is (more exactly, can and will be chosen to be) continuous. The number $L_t(x)$ is called the local time at $x$ of $w$ on $[0, t]$. It is strictly positive for all $x$ of the interval $(a, b)$ where $a$ and $b$ are the least and the greatest value of $w$ on $[0, t]$, respectively. (For $x$ outside this interval the local time evidently vanishes.) Treated as a function of two variables $x$ and $t$, the local time is still continuous. Treated as a function of $t$ (while $x$ is fixed), the local time is a singular function corresponding to a nonatomic measure on the set of zeros of $w$. (from http://en.wikipedia.org/wiki/Wiener_process#Local_time)

Now, I am not getting the following part:

Treated as a function of two variables $x$ and $t$, the local time is still continuous. Treated as a function of $t$ (while $x$ is fixed), the local time is a singular function corresponding to a nonatomic measure on the set of zeros of $w$. (from http://en.wikipedia.org/wiki/Wiener_process#Local_time)

Edit: (OK, I now seem to understand this part: How can local time be a function of $t$? Isn't local time all about measuring time duration of process reaching some particular value $x$? So what does this mean?)

Also, why is it involved with measuring on the set of zeros of $w$? I cannot understand this part.

share|improve this question
    
*How can local time be a function of $t$*... Noticed the small $t$ in $L_t(x)$? –  Did Feb 11 '13 at 14:34
    
I now understand that part. –  Common Knowledge Feb 11 '13 at 14:50

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