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My teacher often use the result How can I prove that $$\overline{\int_{E}f}=\int_{E}\bar f,$$ where $\bar z$ mean the conjugate. But how can I prove it ? Indeed it looks natural since the conjugate of a sum is the sum of the conjugate, and an integral is a sort of continuous sum so I admit that it's true, but I unfortunately can't prove it.

My idea

If $f$ is a simple function, i.e. $f(x)=\sum_{i=1}^n a_i1_{E_i}$ then $$\overline{\int f}=\overline{\sum_{i=1}^n a_i m(E_i)}=\sum_{i=1}^n \bar a_i m(E_i)=\int \bar f.$$

Then, if $f\geq 0$, I take $f_n\nearrow f$ and using the continuity of the conjugate and the previous result, I conclude. To conclude, if $f$ is measurable, the I can right $f=f^+-f^-$ with $f^+,f^-\geq 0$ and conclude as well. Ok it's complicate, so I was wondering if there is an easier method.

user380364
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  • Please remember that you can choose an aswer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 09 '18 at 23:07

4 Answers4

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If $\displaystyle\int f$ exists, then by writing $f=u+iv$ for real $u,v$, then $\displaystyle\int u$ and $\displaystyle\int v$ exist and $\displaystyle\int f=\int u+i\int v$. Note that both $\displaystyle\int u,\int v$ are real, so $\overline{\displaystyle\int f}=\displaystyle\int u-i\int v=\int(u-iv)=\int\overline{f}$.

user284331
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Let $f=g+ih$ then

$$\overline{\int_{E}f}=\overline{\int_{E}g+ih}=\overline{\int_{E}g}+\overline{\int_{E}ih}=\int_{E}g-i\int_{E}h=\int_{E}\bar f,$$

user
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The Lebesgue integral of complex-valued functions is not (that I know of) defined with the same measure-theoretic machinery that you use for $\Bbb R$ (the notions of which, starting from positivity or increasing convergence, would actually be moot if $f$ could take non-real values): it's just defined as the component-wise Lebesgue integral of $f$ as a function $X\to\Bbb C=\Bbb R^2$. The result is an element of $(a,b)\in\Bbb R^2$, which is then identified with the number $a+ib\in\Bbb C$. So, the identity is indeed obvious: you are saying that if $(a,b)=(\int f_1, \int f_2)$, then $(a,-b)=(\int f_1,\int -f_2)$.

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If $f:\mathbb C\to \mathbb R$ it's obvious since the integral of a real function is real. Suppose $f:\mathbb C\to\mathbb C$ and write $$f(z)=u(x,y)+iv(x,y)$$ with $x,y\in\mathbb R$, $u=\Re(f)$ and $v=\Im(f)$. Since $u(x,y),v(x,y)\in\mathbb R$ for all $x,y$, $$\overline{\int f}=\overline{\int u+i\int v}=\overline{\int u}-i\overline{\int v}=\int u-i\int v=\int(u-iv)=\int \bar f.$$

Surb
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