Real Analysis – Linear Combinations of Uniformly Integrable Functions

real-analysis

Let $\{f_n\}$ and $\{g_n\}$ be sequences of uniformly integrable functions on $E$. Show that for $\alpha$, $\beta$, the sequences $\alpha f_n + \beta g_n$ are also uniformly integrable.

Attempt at a proof:

Since both sequences are uniformly integrable I can find a $\delta=\min\{\delta_{g_n}, \delta_{f_n}\}$ such that $|\int_{A_{f_n}\cup A_{g_n}} f_n +g_n|< \epsilon$.

I would like to know how to show this if I knew that $f_n$ would converge pointwise to $f$. I'm not given this detail however.

Edit:

A family $\mathscr{F}$ of measurable functions on $E$ is said to be uniformly integrable over $E$ provided for each $\epsilon >0$, there is a $\delta >0$ such that for each $f \in \mathscr{F}$, if $A \subseteq E$ is measurable and $m(A)<\delta$, the $\int_A |f|< \epsilon$.

Best Answer

The solution depends on the definition of uniform integrability you use. The most convenient one states that:

A collection $H$ of integrable functions bounded in $L^1(\Omega,\mathcal F,\mu)$ is uniformly integrable if $\int\limits_A|h|\to0$ when $\mu(A)\to0$, uniformly on $h$ in $H$, that is, $$ \forall\varepsilon\gt0,\ \exists\eta\gt0,\ \forall A\in\mathcal F,\ \forall h\in H,\ \mu(A)\leqslant\eta\implies\int_A|h|\mathrm d\mu\leqslant\varepsilon. $$

Then the proof is direct. First, if $\alpha H=\{\alpha h\mid h\in H\}$ and $H$ is uniformly integrable, then $\alpha H$ is bounded in $L^1$ and, if $\eta(\varepsilon,H)$ makes the implication above true for $H$ and $\varepsilon$, for each $\varepsilon$, then $\eta\left(\frac{\varepsilon}{|\alpha|},H\right)$ makes it true for $\alpha H$ and $\varepsilon$.

Second, if $H+K=\{h+k\mid h\in H,\ k\in K\}$ and $H$ and $K$ are uniformly integrable, then $H+K$ is bounded in $L^1$ and $\min\{\eta(\varepsilon,H),\eta(\varepsilon,K)\}$ makes the implication above true for $H+K$ and $2\varepsilon$.

Hence for every uniformly integrable collections $H$ and $K$ and every scalar $\alpha$ and $\beta$, $\alpha H+\beta K$ is uniformly integrable as well.

Note finally that the sequences $(f_n)_{n\in\mathbb N}$ and $(g_n)_{n\in\mathbb N}$ are uniformly integrable if and only if the collections $H=\{f_n\mid n\in\mathbb N\}$ and $K=\{g_n\mid n\in\mathbb N\}$ are uniformly integrable and that, then, $\alpha H+\beta K$ contains every function $\alpha f_n+\beta g_n$ (and many more). Hence the uniform integrability of $\alpha H+\beta K$ implies the one of $\{\alpha f_n+\beta g_n\mid n\in\mathbb N\}$.

If one wishes to use the definition that:

A collection $H$ is uniformly integrable if and only if $\int\limits_{|h|\gt t}|h|\mathrm d\mu\to0$ when $t\to+\infty$, uniformly on $h$ in $H$, that is, $$ \forall\varepsilon\gt0,\ \exists t,\ \forall s,\ \forall h\in H,\ s\geqslant t\implies\int_{|h|\geqslant s}|h|\mathrm d\mu\leqslant\varepsilon, $$

then the proof is similar but perhaps less convenient. Denote by $t(\varepsilon,H)$ any value of $t$ making the implication above true for $H$ and $\varepsilon$. Then $t(\varepsilon,H)$ works for $\alpha H$ and $|\alpha|\varepsilon$ hence $H$ uniformly integrable implies $\alpha H$ uniformly integrable. For the sum $H+K$, one can convince oneself of the validity of the pointwise inequality $$ |h+k|\,\mathbf 1_{|h+k|\geqslant 2t}\leqslant 2|h|\,\mathbf 1_{|h|\geqslant t}+2|k|\,\mathbf 1_{|k|\geqslant t}, $$ and deduce that, in our notations, $\max\{t(\varepsilon,H),t(\varepsilon,K)\}$ works for $H+K$ and $4\varepsilon$. QED.

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