integral of f(x,y) w.r.t. x as function of y continuous?

nicster2009 · **Joined:** Wed, 15 Feb 2012 18:56:18 UTC **Posts:** 4

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!

Shadow · **Posted:** Wed, 15 Feb 2012 19:15:52 UTC

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 $f(x,y)=\begin{cases} \infty & x=1 \\ x & x\ne 1\end{cases}$

nicster2009 · **Joined:** Wed, 15 Feb 2012 18:56:18 UTC **Posts:** 4

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.

outermeasure · **Posted:** Thu, 16 Feb 2012 04:11:36 UTC

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.

nicster2009 · **Joined:** Wed, 15 Feb 2012 18:56:18 UTC **Posts:** 4

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

Shadow · **Posted:** Thu, 16 Feb 2012 21:24:23 UTC

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

is ctns a.e., it's equivalence class in L^p

contains a continuous function, so you may as well assume

is continuous for the purpose of integration.

outermeasure · **Posted:** Fri, 17 Feb 2012 03:25:31 UTC

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

. 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

is ctns a.e., it's equivalence class in L^p

contains a continuous function, so you may as well assume

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.

Shadow · **Posted:** Fri, 17 Feb 2012 04:49:43 UTC

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

. 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

is ctns a.e., it's equivalence class in L^p

contains a continuous function, so you may as well assume

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!

nicster2009 · **Joined:** Wed, 15 Feb 2012 18:56:18 UTC **Posts:** 4

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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integral of f(x,y) w.r.t. x as function of y continuous?

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