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Say $f,g$ are two measurable functions $(X,\Sigma)\to(\mathbb{R}^n,\mathcal{L}(\mathbb{R}^n))$. Are $f+g$ and $rf$ still measurable functions? Here $r\in\mathbb{R}$.

So the motivation of this question is: if $f,g\in L^1(\mathbb{R}^n)$, then for almost every $x\in\mathbb{R}^n$, the convolution $$(f*g)(x):=\int_{\mathbb{R}^n}f(y)g(x-y)\,dm(y)$$ exists, and $$\|f*g\|_1\leq\|f\|_1\|g\|_1.$$ The proof is not hard, just an application of Fubini's Theorem. But first of all, we need to show the function $(x,y)\mapsto f(y)g(x-y)$ is measurable from $(\mathbb{R}^n\times\mathbb{R}^n,\mathcal{L}(\mathbb{R}^n)\times\mathcal{L}(\mathbb{R}^n))$ to $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$. I believe proving this fact requires us to prove what I asked above.

Zhulin Li
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  • This is tricky, usually the other side of the map is a Borel set. Or the normal definition of measurable function should be to inverse-map every Borel set of the image back to a measurable set. – Jay Zha Mar 24 '17 at 01:04
  • Any continuous function is universally measurable; and sum and multiplication by scalar are continuous. Am I missing some point? – William M. Mar 24 '17 at 01:16
  • Can you write more clearly what are you asking for? The standard definition of a "real valued function" to be "measurable" is using the Borel sigma algebra. Seems strange that you want the target to be Lebesgue sigma algebra. I have never seen that. – William M. Mar 24 '17 at 03:16
  • For instance, a function $f:\Bbb R \to \Bbb R$ to be measurable means $f^{-1}(B)$ is a Lebesgue set for every Borel set $B.$ In this case, you can do all the operations because all operations are continuous (hence, Borel-Borel measurable). – William M. Mar 24 '17 at 03:17
  • Related; there it is proved that $(x,y) \mapsto g(x-y)$ is measurable ($(\mathbb{R}^{2n},\mathcal{L}(\mathbb{R}^{2n})) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$, then concluding the same for codomain $\mathbb{R}^k$ with its Borel $\sigma$-algebra is trivial). It boils down to showing $(x,y) \mapsto x-y$ is $\mathcal{L}(\mathbb{R}^{2n}) \to \mathcal{L}(\mathbb{R}^n)$-measurable. The same proof with small modifications works for addition rather than subtraction of course, so that – Daniel Fischer Mar 24 '17 at 15:46
  • also shows the $\Sigma \to \mathcal{L}(\mathbb{R}^n)$-measurability of $f+g$. The $rf$ part is much easier, since either $r = 0$, whence $rf$ is constant, so measurable, or $r\neq 0$ and then $x\mapsto rx$ completely preserves Lebesgue-measurability of subsets ($S$ is Lebesgue-measurable if and only if $r\cdot S$ is). – Daniel Fischer Mar 24 '17 at 15:46
  • @WillM. So I am working in abstract measure theory, not just real analysis. Say $(X,\mathcal{X})$ and $(Y,\mathcal{Y})$ are two measurable spaces. Then we say a function $f:X\to Y$ is measurable if $f^{-1}(B)\in\mathcal{X}$ for all $B\in\mathcal{Y}$. Now here is my question. Let $(X,\mathcal{X})$ be a measurable space. Let $f,g$ be two measurable functions from $(X,\mathcal{X})$ to $(\mathbb{R^n},\mathcal{L}(\mathbb{R^n}))$. Is $f+g$ still measurable from $(X,\mathcal{X})$ to $(\mathbb{R^n},\mathcal{L}(\mathbb{R^n}))$? – Zhulin Li Mar 25 '17 at 03:23
  • Possible duplicate of http://math.stackexchange.com/questions/541118/proving-that-sum-of-two-measurable-functions-is-measurable – Adam Mar 25 '17 at 05:10
  • @Adam It is not a duplicate because, as my previous comment pointed out, the standard definition of measurable is that of the target space having Borel sigma algebra. OP is just wondering if sum of two functions is still measurable when the target has Lebesgue sigma field. I am not quite sure, actually, about this totally different statement. My first guess would be that it is false. – William M. Mar 26 '17 at 04:29

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