Prove that $m(\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\})\geq\frac{1}{2}$ under these conditions...

Question

Question: Suppose for each $j\in\mathbb{N}, f_j:[0,1]\rightarrow\mathbb{R}$ is Lebesgue measurable such that $0\leq f_j\leq\frac{3}{2}$ and $\int_0^1 f_j dm=1$. Prove that $m(\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\})\geq\frac{1}{2}$.

Thoughts/Attempt: Let $A=\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\}$, and $B=\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)<\frac{1}{2}\}$. Suppose, by contradiction, that $m(A)<\frac{1}{2}$. So, we can split up the integral as $$\int_0^1f_jdm=\int_Af_jdm+\int_Bf_jdm$$ where we get equality by the integral in the assumption. Now, $\int_Af_jdm<\frac{1}{2}$, by our (contradiction) assumption. And, $\int_Bf_jdm<\frac{1}{2}$, using our set $B$. Therefore, $$\int_0^1f_jdm<\frac{1}{2}+\frac{1}{2}=1$$ a contradiction, since this integral must equal $1$ from our assumption. Hence, we contradict that $m(A)<\frac{1}{2}$.

However, I am not quite sure if this works, because our sets are dealing with the $\lim\sup f_j(x)$ as $x\in[0,1]$, but wouldn't I have to compensate in the integral since the image of $f_j$ is all of $\mathbb{R}$?

Answer 1

To handle the $\limsup$ for your integrals over $B$, I suggest using Fatou's lemma.

Fatou's lemma only applies to non-negative functions, and $\tfrac 3 2 - f_n(x)$ is a non-negative function. Applying Fatou to $\tfrac 3 2 - f_n(x)$ on $B$, we have $$ \int_B \liminf_{n \to \infty}\left(\tfrac 3 2 - f_n (x)\right) dm \leq \liminf_{n \to \infty} \int_B \left(\tfrac 3 2 - f_n (x)\right) dm,$$ or equivalently, $$ \limsup_{n \to \infty} \int_B f_n(x) dm \leq \int_B \limsup_{n \to \infty} f_n(x) \leq \tfrac 1 2 m(B).$$

Meanwhile, the correct inequality for the integrals over $A$ is $$ \int_A f_n(x) dm \leq \tfrac 3 2 m(A)$$ for each $n \in \mathbb N$. (It seems like you've missed the factor of $\tfrac 3 2$, coming from the upper bound on $f_n$.)

Thus $$ 1 = \limsup_{n \to \infty}\int_0^1 f_n(x) dx \leq \tfrac 3 2 m(A) + \tfrac 1 2 m(B) = m(A) + \tfrac 1 2,$$ which clearly implies that $m(A) \geq \tfrac 1 2$.

Answer 2

For each $n\in\mathbb{N}$ $$\begin{align} 1=\int f_n &=\int_{\{f_n\geq\frac12\}}f_n+\int_{\{f_n<\frac12\}}f_n\leq \frac32\lambda\Big(f_n\geq\frac12\Big) + \frac12\lambda\Big(f_n<\frac12\Big)\\ &=\lambda\Big(f_n\geq\frac12\Big)+\frac12 \end{align}$$ Hence $$\lambda\Big(f_n\geq\frac12\Big)\geq\frac12\qquad\text{for all}\quad n\in\mathbb{N}$$

Notice that $$\begin{align} \Big\{\limsup_jf_j\geq\frac12\Big\}\supset\limsup_j\Big\{f_j\geq\frac12\Big\}=\bigcap_n\bigcup_{m\geq n}\Big\{f_m\geq\frac12\Big\}\tag{1}\label{one} \end{align}$$

Putting things together we obtain $$\lambda\Big(\limsup_jf_j\geq\frac12\Big)\geq\lim_n\lambda\Big(\bigcup_{m\geq n}\{f_m\geq\frac12\}\Big)\geq\limsup_n\lambda\Big(f_n\geq\frac12\Big)\geq\frac12$$

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Comment:

The set inequality $\eqref{one}$ is clear since $x\in\bigcap_n\bigcup_{m\geq n}\Big\{f_n\geq\frac12\Big\}$ implies that $f_n(x)\geq\frac12$ infinitely many times, and so $\limsup_nf_n(x)\geq\frac12$.