# Proof verification: Show $f_n\to f$ in measure given $\int f_n \to \int \limsup f_n$ on $[0,1]$

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?

• What does $m(x | \int f - f_n > \delta - \epsilon)$ even mean? – James Yang Nov 4 at 4:31
• 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. – James Yang Nov 4 at 4:39