Convergence in Sigma-Finite Measure
Jump to navigation
Jump to search
Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space.
Let $\left({f_n}\right)_{n \in \N}, f_n: X \to \R$ be a sequence of measurable functions.
Also, let $f, g: X \to \R$ be measurable functions.
Suppose that $f_n$ converges in measure to both $f$ and $g$ (in $\mu$).
Then $f$ and $g$ are equal $\mu$-almost everywhere.
Proof
This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $16.5$