MHB How is √2 μ Interpreted Geometrically?

[Juliayaho]

This is part 2 of a question... I already solved part 1 but I can't seem to be able to solve this one.
Interpret the measure √2 μ geometrically?

Any ideas... This is from real analysis class

Thanks in advance!

 

[Prove It]

Re: Measure... Geometrically?

This is part 2 of a question... I already solved part 1 but I can't seem to be able to solve this one.
Interpret the measure √2 μ geometrically?

Any ideas... This is from real analysis class

Thanks in advance!

What is [math]\displaystyle \mu [/math] representing?

 

[Juliayaho]

Re: Measure... Geometrically?

The unique positive regular measure on B(R^2) such that for all f in C_c(R^2) λf= ∫_R^2 f dµ

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What is [math]\displaystyle \mu [/math] representing?

The unique positive regular measure on B(R^2) such that for all f in C_c(R^2) λf= ∫_R^2 f dµ

- - - Updated - - -

The unique positive regular measure on B(R^2) such that for all f in C_c(R^2) λf= ∫_R^2 f dµ

- - - Updated - - -
The unique positive regular measure on B(R^2) such that for all f in C_c(R^2) λf= ∫_R^2 f dµ

λf=∫(from R to Riemann) f(tt)dt for all f in C_c(R^2).

 

[Ackbach]

Re: Measure... Geometrically?

I could be wrong, but it seems to me that this question could be answered much more easily (that is, intuitively) by thinking about the corresponding Riemann integral. It's a fact that if the function is Riemann integrable, then
$$\int_{ \mathbb{R}^{2}} f \, dA = \int_{ \mathbb{R}^{2}} f \, d\mu.$$
So take an example, and make $f$ really simple, say $f=1$. Then your integral is simply computing area. Your new measure, $\sqrt{2} \mu$, simply multiplies your area by $\sqrt{2}$, since constants pull out of integrals (at least, they pull out of Riemann and Lebesgue integrals). What is going on geometrically for that to happen?