Let $(\mathbb{R},\mathscr{P}(\mathbb{R}),\phi)$ be a measure space. Let $A$ be a subset of $\mathbb{R}$ such that $\phi (A) > 0$, $\phi (A^c)= 0$ and $\mathbf{card(A)=n<\infty}$. Then $\mathcal L ^{1}(\phi)=\mathcal L ^{2}(\phi)$. $\hspace{5mm}$Is my proof correct? It is as follows:
($\mathcal L ^{p}(\phi)= \{f:\mathbb{R} \rightarrow \mathbb{C} : f$ is measurable$,\int_{\mathbb{R}}|f|^p <\infty \}$)
Let $f \in \mathcal L ^{1}(\phi)$. Since $\phi (A^c)= 0$ and $\mathcal{X}_{\{x\}}$ is measurable $\forall x \in \mathbb{R}$ ($\{x\}\in \mathscr{P}(\mathbb{R}), \forall x \in \mathbb{R}$) we have that: $\int_\mathbb{R} |f| \hspace{1mm} d\phi =\int_A |f| \hspace{1mm} d\phi +\int_{A^c} |f| \hspace{1mm} d\phi =\int_A |f| \hspace{1mm} d\phi=\int_A |f|_{|_{A}} \hspace{1mm} d\phi=\int_A \sum_{x\in A} |f(x)|\mathcal{X}_{\{x\}}\hspace{1mm} d\phi = \sum_{x\in A} \int_A |f(x)|\mathcal{X}_{\{x\}}\hspace{1mm} d\phi=\sum_{x\in A} \int_{\{x\}} |f(x)|\hspace{1mm} d\phi = \sum_{x\in A} |f(x)|\hspace{1mm}\phi (\{x\})<\infty$.
It is immediate that $\forall x\in A$ such that $\phi (x)= \infty $ then we have that $|f(x)|=0$ (If we supposed the contrary we would have that $\int_\mathbb{R} |f| \hspace{1mm} d\phi = \infty$). Hence $|f(x)|^2 = 0$, $ \forall x\in A $ such that $ \phi(\{x\})=\infty$.
Then, we conclude that $\int_\mathbb{R} |f|^2 \hspace{1mm} d\phi= \sum_{x\in A} |f(x)|^2\hspace{1mm}\phi (\{x\})< \infty$. Then $\mathcal L ^{1}(\phi)\subset\mathcal L ^{2}(\phi)$.
To prove that $\mathcal L ^{2}(\phi)\subset\mathcal L ^{1}(\phi)$ we carry out the same reasoning.