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PostPosted: Wed, 15 Feb 2012 19:05:33 UTC 
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Hi, my intuition tells me that \int f(x,y)dy should be a continuous function of x if f(x,y) is continuous almost everywhere. I'm not asking that someone should prove this for me (if possible), but I'm wondering whether anyone could point me towards a reference where this is shown. I have been searching and not finding anything. Thanks!


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PostPosted: Wed, 15 Feb 2012 19:15:52 UTC 
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nicster2009 wrote:
Hi, my intuition tells me that \int f(x,y)dy should be a continuous function of x if f(x,y) is continuous almost everywhere. I'm not asking that someone should prove this for me (if possible), but I'm wondering whether anyone could point me towards a reference where this is shown. I have been searching and not finding anything. Thanks!


No, let f(x,y)=\begin{cases} \infty & x=1 \\ x & x\ne 1\end{cases}

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PostPosted: Wed, 15 Feb 2012 23:58:53 UTC 
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Thanks. But what if f is bounded? If it makes a difference, in my case the domains of x and y are closed intervals and f is real valued.


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PostPosted: Thu, 16 Feb 2012 04:11:36 UTC 
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nicster2009 wrote:
Thanks. But what if f is bounded? If it makes a difference, in my case the domains of x and y are closed intervals and f is real valued.


Assuming finiteness of the measure you put on the closed interval (on which y is allowed to reside), it is just a matter of using uniform continuity of f.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Thu, 16 Feb 2012 12:20:16 UTC 
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But I don't even assume that f is continuous...


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PostPosted: Thu, 16 Feb 2012 21:24:23 UTC 
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nicster2009 wrote:
But I don't even assume that f is continuous...


You have that L^\infty[a,b] is in L^p[a,b] for p\ge 1, then since f is ctns a.e., it's equivalence class in L^p contains a continuous function, so you may as well assume f is continuous for the purpose of integration.

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PostPosted: Fri, 17 Feb 2012 03:25:31 UTC 
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nicster2009 wrote:
But I don't even assume that f is continuous...


Oops, sorry misread the question.

The answer is no. For example, let f(x,y)=1_A(x) for some A. Then a.e. continuity of f is equivalent to \partial A is null. But then \int_I f(x,y)\,\mathrm{d}y=\lvert I\rvert\cdot 1_A(x) is not continuous unless A is ...

On the other hand, you have \int f(x,y)\,\mathrm{d}y is continuous almost everywhere. (Try to prove it!)

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nicster2009 wrote:
But I don't even assume that f is continuous...


You have that L^\infty[a,b] is in L^p[a,b] for p\ge 1, then since f is ctns a.e., it's equivalence class in L^p contains a continuous function, so you may as well assume f is continuous for the purpose of integration.


Actually no. For example, a step function is continuous almost everywhere (since there is only discontinuous at the end-point of the interval), but admits no continuous representative.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 17 Feb 2012 04:49:43 UTC 
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outermeasure wrote:
nicster2009 wrote:
But I don't even assume that f is continuous...


Oops, sorry misread the question.

The answer is no. For example, let f(x,y)=1_A(x) for some A. Then a.e. continuity of f is equivalent to \partial A is null. But then \int_I f(x,y)\,\mathrm{d}y=\lvert I\rvert\cdot 1_A(x) is not continuous unless A is ...

On the other hand, you have \int f(x,y)\,\mathrm{d}y is continuous almost everywhere. (Try to prove it!)

Shadow wrote:
nicster2009 wrote:
But I don't even assume that f is continuous...


You have that L^\infty[a,b] is in L^p[a,b] for p\ge 1, then since f is ctns a.e., it's equivalence class in L^p contains a continuous function, so you may as well assume f is continuous for the purpose of integration.


Actually no. For example, a step function is continuous almost everywhere (since there is only discontinuous at the end-point of the interval), but admits no continuous representative.


You're right, and I see exactly where I was off in my thinking, thanks for that outermeasure!

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PostPosted: Fri, 17 Feb 2012 19:14:39 UTC 
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Thanks a lot. I had been working on a more restricted version of the same problem a while ago and from that I took away that it should hold. The counter example is quite straightforward though. I should have thought a bit more about the general case before posting.


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