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Suppose that $f+g$ is a continuous functions on $[a,b]$, then $\mu(\Gamma(f+g))=0$. Here we are considering the Lebesgue measure in $\mathbb{R}^2$. Additionally, $\Gamma(f)$ represents the graph of the function $f$.

I was interested in whether we can say anything about $\mu(\Gamma(f))$ and $\mu(\Gamma(g))$ here? Could I it be possible that $\mu(\Gamma(f))=0$ and $\mu(\Gamma(g))=0$ is equivalent to $\mu(\Gamma(f+g))=0$?

JayP
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    $f + (-f) = 0$ is continuous, which means that nothing can be said about $f$. – Martin R Apr 14 '22 at 13:23
  • @MartinR. If we were to impose the condition that $f,g$ are also continuous, this would allow us to say that $\mu(f)=\mu(g)=0$ right? – JayP Apr 14 '22 at 13:41
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    Example of @MartinR still works for any continuous $f$, for example any constant, and hence $\mu(f)$ can be any real value either. – SBF Apr 14 '22 at 14:31
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    What is the meaning of $\mu(f+g)$, $\mu(f)$, and $\mu(g)$? – Michael Apr 14 '22 at 14:33
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    @Michael in measure theory it is common to treat measures (functions of sets) as functions of functions using the Lebesgue integral construction. That is, $\mu(f) = \int f\mathrm d\mu$ and in particular $\mu(1_A) = \mu(A)$ for indicator functions. – SBF Apr 14 '22 at 15:06
  • @Ilya : I doubt that is what the asker MATHBOI means. It would not be consistent with using the Lebesgue measure in $\mathbb{R}^2$, or with the vaguely worded claim in the question that $\mu(h)=0$ if $h$ is a continuous function. I am guessing that $f$ is assumed to be a function $f:\mathbb{R}\rightarrow\mathbb{R}$ and MATHBOI really means $\mu(graph(f))$, meaning the Lebesgue measure in $\mathbb{R}^2$ of the set ${(x, f(x)): x \in \mathbb{R}}$. This illustrates that simple descriptors of the problem are needed. – Michael Apr 14 '22 at 18:45
  • @Michael, you are correct is saying that I am interested in the graph of these functions. I have updated my question accordingly. – JayP Apr 15 '22 at 00:03
  • It seems clear that $\mu(\Gamma(-h))=\mu(\Gamma(h))$. If we assume there are functions $h:\mathbb{R}\rightarrow\mathbb{R}$ such that $\mu(\Gamma(h))\neq 0$ then the very first comment proves you do not have an equivalence. However I believe the forward direction holds: $$\mu(\Gamma(f))=0, \mu(\Gamma(g))=0\implies \mu(\Gamma(f+g))=0$$ – Michael Apr 15 '22 at 02:28
  • @Michael perhaps both of my claims are correct here if both $f,g$ are functions? I found this in another post https://math.stackexchange.com/questions/35606/lebesgue-measure-of-the-graph-of-a-function – JayP Apr 15 '22 at 02:46
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    If both $f$ and $g$ are continuous then $f+g$ is also continuous. – Michael Apr 15 '22 at 02:55

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