We are given that $\int f_n \to \int \limsup{f_n}$

By properties of $\limsup$ we know that $\limsup f_n +\epsilon>f_n$ for large enough $n,$ for each $x$. We will now call $\limsup f_n=f$. Thus $\int f-f_n>-\epsilon$ for big enough $n$ since we are over $[0,1]$. Now by monotonicity we know that $m(x\mid f-f_n>\delta-\epsilon)\leq m(x\mid\int f-f_n>\delta-\epsilon)$. Now whatever $\delta$ is, we can pick $\epsilon<\delta$ so that $\delta-\epsilon$ is positive. Then since $\int f_n \to \int f$we know that $n$ going to infinity. But then $m(x\mid f-f_n>\delta-\epsilon)\to 0$ as $n \to \infty$ this works for any $\delta$ as you can just select $\delta+1$ and $\epsilon=1$ in our proof.

Is this correct?

  • $\begingroup$ What does $m(x | \int f - f_n > \delta - \epsilon)$ even mean? $\endgroup$ – James Yang Nov 4 '19 at 4:31
  • $\begingroup$ Alternatively, we know that $f_n - \sup\limits_{k\geq n} f_k \xrightarrow{L^1} 0$ by the given hypothesis, and by definition of $\limsup$ and MCT, $\sup\limits_{k\geq n} f_k \xrightarrow{L^1} \limsup f_n$, so by linearity, $f_n \xrightarrow{L^1} \limsup f_n$. This immediately implies convergence in measure. $\endgroup$ – James Yang Nov 4 '19 at 4:39

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