0
$\begingroup$

Assume we are given a function space $\mathcal{A} = \{f : \mathcal{X} \mapsto \mathcal{Y}, f \text{ is integrable} \}$. For any $f \in \mathcal{A}$, we define a constant $c = \int_{x \in \mathcal{Z}}f(x)dx$ where $\mathcal{Z} \subseteq \mathcal{X}$ and a function $g : \mathcal{X} \mapsto \mathcal{Y}$ such that for any $x \in \mathcal{X}$, $g(x) = f(x) - \int_{x \in \mathcal{Z}}f(x)dx$. Now it is clear that $f(x) = g(x) +c$. If we define the space$\mathcal{B} = \{c + g(.) : c \text{ is a constant}, g \in \mathcal{C} \}$ where $\mathcal{C} = \{g:\mathcal{X} \mapsto \mathcal{Y}, \int_{x \in \mathcal{Z}}g(x)dx =0 \}$, how can we show that $\mathcal{A} =\mathcal{B}$?

  • 0
    You seem to use $c$ ambiguously. Also, what kind of integral is that? – 2017-02-06
  • 0
    @aduh What I want to show that any $f \in \mathcal{A}$ is equivalent to a constant and a function $g \in \mathcal{B}$. Does it matter what type of integral it is? – 2017-02-06
  • 0
    Just not sure I understand your notation. It seems like the problem is trivial given how you've defined things. By linearity of the integral $\int g = \int f - c = c-c = 0.$ – 2017-02-06
  • 0
    @aduh Yes, it is trivial. I just don't know how to write it formally. – 2017-02-06
  • 0
    In fact, it seems that you've shown $\mathcal{A} \subset \mathcal{B}$. For the reverse inclusion note that the function $x \mapsto g(x) + c$ is integrable if $g$ is, and hence is a member of $\mathcal{A}$. – 2017-02-06
  • 0
    why $\mathcal{A} \subset \mathcal{B}$? – 2017-02-06

1 Answers 1

1

You have already shown that $\mathcal{A} \subset \mathcal{B}$. That is, an integrable function $f \in \mathcal{A}$ can be written as $f = g + c$, where $c = \int f$ and $\int g = 0$. And, hence, $f \in \mathcal{B}$.

For the reverse inclusion, suppose $f \in \mathcal{B}$. Then $f = g+c$ where $\int g = 0$ and $c$ is a constant. Hence, $f$ is integrable and $f \in \mathcal{A}$. This shows $\mathcal{B} \subset \mathcal{A}$.

From both subset inclusions we conclude $\mathcal{A} = \mathcal{B}$.