Mixing
Why does the “$i$” in $f=u+vi$ not affect the measurability of $f$?
I stumbled across the following result in Rudin's Real and Complex Analysis:
He says in part (c) that "the complex case then follows from (a) and (b)." To me, this is saying that because $f$ and $g$ are sums of real-valued functions, i.e. $$f=u_1+iv_1, quad g=u_2+iv_2$$ for real valued $u_1,u_2,v_1$ and $v_2$, the sum of $f$ and $g$ must also be measurable, i.e. $$f+g=
underbrace{(u_1+u_2)}_{text{measurable}}+iunderbrace{(v_1+v_2)}_{text{measurable}}$$
What I can't seem to get past is how the $i$ was seemingly ignored! Why are we allowed to ignore it?
real-analysis complex-analysis measure-theory proof-explanation
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
add a comment |
I stumbled across the following result in Rudin's Real and Complex Analysis:
He says in part (c) that "the complex case then follows from (a) and (b)." To me, this is saying that because $f$ and $g$ are sums of real-valued functions, i.e. $$f=u_1+iv_1, quad g=u_2+iv_2$$ for real valued $u_1,u_2,v_1$ and $v_2$, the sum of $f$ and $g$ must also be measurable, i.e. $$f+g=
underbrace{(u_1+u_2)}_{text{measurable}}+iunderbrace{(v_1+v_2)}_{text{measurable}}$$
What I can't seem to get past is how the $i$ was seemingly ignored! Why are we allowed to ignore it?
real-analysis complex-analysis measure-theory proof-explanation
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
5
Read part (a) again.
– Eric Wofsey
Nov 21 '18 at 3:14
1
Do you know how $mathbf{C}$ is defined?
– Will M.
Nov 21 '18 at 3:57
@EricWofsey After reading (a) numerous times, it made sense with MonstrousMoonshiner's answer. Thank you for the nudge in the right direction! And Will M., I know that $mathbf{C}={a+bi: a,binmathbf{R}}$.
– Thy Art is Math
Nov 21 '18 at 4:07
add a comment |
I stumbled across the following result in Rudin's Real and Complex Analysis:
He says in part (c) that "the complex case then follows from (a) and (b)." To me, this is saying that because $f$ and $g$ are sums of real-valued functions, i.e. $$f=u_1+iv_1, quad g=u_2+iv_2$$ for real valued $u_1,u_2,v_1$ and $v_2$, the sum of $f$ and $g$ must also be measurable, i.e. $$f+g=
underbrace{(u_1+u_2)}_{text{measurable}}+iunderbrace{(v_1+v_2)}_{text{measurable}}$$
What I can't seem to get past is how the $i$ was seemingly ignored! Why are we allowed to ignore it?
real-analysis complex-analysis measure-theory proof-explanation
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
I stumbled across the following result in Rudin's Real and Complex Analysis:
He says in part (c) that "the complex case then follows from (a) and (b)." To me, this is saying that because $f$ and $g$ are sums of real-valued functions, i.e. $$f=u_1+iv_1, quad g=u_2+iv_2$$ for real valued $u_1,u_2,v_1$ and $v_2$, the sum of $f$ and $g$ must also be measurable, i.e. $$f+g=
underbrace{(u_1+u_2)}_{text{measurable}}+iunderbrace{(v_1+v_2)}_{text{measurable}}$$
What I can't seem to get past is how the $i$ was seemingly ignored! Why are we allowed to ignore it?
real-analysis complex-analysis measure-theory proof-explanation
real-analysis complex-analysis measure-theory proof-explanation
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
asked Nov 21 '18 at 3:03
Thy Art is Math
489211
489211
5
Read part (a) again.
– Eric Wofsey
Nov 21 '18 at 3:14
1
Do you know how $mathbf{C}$ is defined?
– Will M.
Nov 21 '18 at 3:57
@EricWofsey After reading (a) numerous times, it made sense with MonstrousMoonshiner's answer. Thank you for the nudge in the right direction! And Will M., I know that $mathbf{C}={a+bi: a,binmathbf{R}}$.
– Thy Art is Math
Nov 21 '18 at 4:07
add a comment |
5
Read part (a) again.
– Eric Wofsey
Nov 21 '18 at 3:14
1
Do you know how $mathbf{C}$ is defined?
– Will M.
Nov 21 '18 at 3:57
@EricWofsey After reading (a) numerous times, it made sense with MonstrousMoonshiner's answer. Thank you for the nudge in the right direction! And Will M., I know that $mathbf{C}={a+bi: a,binmathbf{R}}$.
– Thy Art is Math
Nov 21 '18 at 4:07
5
5
Read part (a) again.
– Eric Wofsey
Nov 21 '18 at 3:14
Read part (a) again.
– Eric Wofsey
Nov 21 '18 at 3:14
1
1
Do you know how $mathbf{C}$ is defined?
– Will M.
Nov 21 '18 at 3:57
Do you know how $mathbf{C}$ is defined?
– Will M.
Nov 21 '18 at 3:57
@EricWofsey After reading (a) numerous times, it made sense with MonstrousMoonshiner's answer. Thank you for the nudge in the right direction! And Will M., I know that $mathbf{C}={a+bi: a,binmathbf{R}}$.
– Thy Art is Math
Nov 21 '18 at 4:07
@EricWofsey After reading (a) numerous times, it made sense with MonstrousMoonshiner's answer. Thank you for the nudge in the right direction! And Will M., I know that $mathbf{C}={a+bi: a,binmathbf{R}}$.
– Thy Art is Math
Nov 21 '18 at 4:07
add a comment |
1 Answer
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Per part (b), $u_1, u_2, v_1$, and $v_2$ are real-measurable functions on $X$, and thus $u_1 + u_2$ and $v_1 + v_2$ are real-measurable functions on $X$. Per part (a), this implies that $f+g = (u_1 + u_2) + i(v_1 + v_2)$ is a complex-measurable function on $X$. In particular, the $i$ is baked in to the statement of part (a).
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Per part (b), $u_1, u_2, v_1$, and $v_2$ are real-measurable functions on $X$, and thus $u_1 + u_2$ and $v_1 + v_2$ are real-measurable functions on $X$. Per part (a), this implies that $f+g = (u_1 + u_2) + i(v_1 + v_2)$ is a complex-measurable function on $X$. In particular, the $i$ is baked in to the statement of part (a).
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
add a comment |
Per part (b), $u_1, u_2, v_1$, and $v_2$ are real-measurable functions on $X$, and thus $u_1 + u_2$ and $v_1 + v_2$ are real-measurable functions on $X$. Per part (a), this implies that $f+g = (u_1 + u_2) + i(v_1 + v_2)$ is a complex-measurable function on $X$. In particular, the $i$ is baked in to the statement of part (a).
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
add a comment |
Per part (b), $u_1, u_2, v_1$, and $v_2$ are real-measurable functions on $X$, and thus $u_1 + u_2$ and $v_1 + v_2$ are real-measurable functions on $X$. Per part (a), this implies that $f+g = (u_1 + u_2) + i(v_1 + v_2)$ is a complex-measurable function on $X$. In particular, the $i$ is baked in to the statement of part (a).
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
Per part (b), $u_1, u_2, v_1$, and $v_2$ are real-measurable functions on $X$, and thus $u_1 + u_2$ and $v_1 + v_2$ are real-measurable functions on $X$. Per part (a), this implies that $f+g = (u_1 + u_2) + i(v_1 + v_2)$ is a complex-measurable function on $X$. In particular, the $i$ is baked in to the statement of part (a).
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
answered Nov 21 '18 at 3:51
Monstrous Moonshiner
2,25011337
2,25011337
add a comment |
add a comment |
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5
Read part (a) again.
– Eric Wofsey
Nov 21 '18 at 3:14
1
Do you know how $mathbf{C}$ is defined?
– Will M.
Nov 21 '18 at 3:57
@EricWofsey After reading (a) numerous times, it made sense with MonstrousMoonshiner's answer. Thank you for the nudge in the right direction! And Will M., I know that $mathbf{C}={a+bi: a,binmathbf{R}}$.
– Thy Art is Math
Nov 21 '18 at 4:07