Calculating the integral in measure theory

Question

I'm very new to measure theory, so I don't really understand how to do this: $\int_Axe^yd\lambda^2$ for A:=$\left\{(x,y)\in R : 0\le x\le 1, 0\le y\le \frac1 2 x^2\right\}$

Do I calculate it using regular integral rules? I'm sorry if this sounds like a I don't know anything question, but I am truly lost and I hope that someone can still help out, thanks in advance!

score 1 · Accepted Answer · answered Jan 31 '23 at 20:40

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Assuming $\lambda^2$ is supposed to be the Lebesgue measure on $\mathbb R^2$, you can simply treat this integral as a Riemann integral, since $(x,y)\mapsto x\cdot e^y$ is continuous and therefore Riemann integrable. In this case, the Riemann integral and Lebesgue integral are equal, see here.

Now, the Riemann integral can be written as a double integral of the form $$\int_ *^*\int_*^*xe^y dxdy$$ You'll just have to figure out the integration boundaries $(*)$.

answered Jan 31 '23 at 20:40

Small Deviation

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Would it be correct that the first integral we have [0,1] and the second has [0, 1/2] boundrie? – ugjumb Feb 01 '23 at 19:51
@ugjumb Not quite. Note that $x$ ranges from $0$ to $1$ and $y$ ranges from $0$ to $\frac{1}{2}x^2$, so at least one of the integration boundaries needs to have an "$x$" in it. Also, try integrating w.r.t. to $y$ first, then w.r.t. $x$, meanin $\int\int xe^y dydx$ – Small Deviation Feb 01 '23 at 19:58
Really? because if we take into account that x can be max 1, so won't that mean that y can only go to 1/2 oor does that train of thought make no sense? – ugjumb Feb 01 '23 at 20:04
Doing the dy first I got xe-e, then without boundries dx $e(\frac{x^2} 2 - x)$ – ugjumb Feb 01 '23 at 20:06
$y$ can go at most to $1/2$, but that does not mean that $y$ $\textbf{ does }$ go up to $1/2$. It only goes to $1/2x^2$ for fixed $x$. – Small Deviation Feb 01 '23 at 20:10
So it would be fine to put x2\2 as boundry? – ugjumb Feb 01 '23 at 20:14
Yes, assuming you mean $x^2/2$. – Small Deviation Feb 01 '23 at 20:23
I GOT IT! THANK YOU!!!!!! – ugjumb Feb 01 '23 at 21:04

score 0 · Answer 2 · answered Jan 31 '23 at 21:19

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Here is a theorem that you might find handy:

Let $f \colon \varOmega \to \mathbb R$ be measurable and $f \geq 0$ almost everywhere. Then \begin{equation} \int f \, d\lambda = \int_0^{\infty} \lambda\left(\{f \geq t\}\right) \, dt \end{equation}

answered Jan 31 '23 at 21:19

Novice

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Calculating the integral in measure theory

2 Answers2