Fatou's lemma

This proof does not rely on the monotone convergence theorem. However, we do explain how that theorem may be applied.

For those not interested in independent proof, the intermediate results below may be skipped.

Lemma 1. Let ${\displaystyle (\Omega ,{\mathcal {F}},\mu )}$ be a measurable space. Consider a simple ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable non-negative function ${\displaystyle s:\Omega \to {\mathbb {R} _{\geq 0}}}$. For a subset ${\displaystyle S\subseteq \Omega }$, define

${\displaystyle \nu (S)=\int _{S}s\,d\mu }$.

Then ${\displaystyle \nu }$ is a measure on ${\displaystyle \Omega }$.

We will only prove countable additivity, leaving the rest up to the reader. Let ${\displaystyle S=\cup _{i=1}^{\infty }S_{i}}$, where all the sets ${\displaystyle S_{i}}$ are pairwise disjoint. Due to simplicity,

${\displaystyle s=\sum _{i=1}^{n}c_{i}\cdot {\mathbf {1} }_{A_{i}}}$,

for some finite non-negative constants ${\displaystyle c_{i}\in {\mathbb {R} }_{\geq 0}}$ and pairwise disjoint sets ${\displaystyle A_{i}\in {\mathcal {F}}}$ such that ${\displaystyle \cup _{i=1}^{n}A_{i}=\Omega }$. By definition of Lebesgue integral,

${\displaystyle {\begin{aligned}\nu (S)&=\\&=\sum _{i=1}^{n}c_{i}\cdot \mu (S\cap A_{i})\\&=\sum _{i=1}^{n}c_{i}\cdot \mu {\bigl (}(\cup _{j=1}^{\infty }S_{j})\cap A_{i}{\bigr )}\\&=\sum _{i=1}^{n}c_{i}\cdot \mu {\bigl (}\cup _{j=1}^{\infty }(S_{j}\cap A_{i}){\bigr )}\end{aligned}}}$

Since all the sets ${\displaystyle S_{j}\cap A_{i}}$ are pairwise disjoint, the countable additivity of ${\displaystyle \mu }$ gives us

${\displaystyle \sum _{i=1}^{n}c_{i}\cdot \mu {\bigl (}\cup _{j=1}^{\infty }(S_{j}\cap A_{i}){\bigr )}=\sum _{i=1}^{n}c_{i}\cdot \sum _{j=1}^{\infty }\mu (S_{j}\cap A_{i}).}$

Since all the summands are non-negative, the sum of the series, whether this sum is finite or infinite, cannot change if summation order does because the series is either absolutely convergent or diverges to ${\displaystyle +\infty .}$ For that reason,

${\displaystyle {\begin{aligned}\sum _{i=1}^{n}c_{i}\cdot \sum _{j=1}^{\infty }\mu (S_{j}\cap A_{i})&=\sum _{j=1}^{\infty }\sum _{i=1}^{n}c_{i}\cdot \mu (S_{j}\cap A_{i})\\&=\sum _{j=1}^{\infty }\int _{S_{j}}s\,d\mu \\&=\sum _{j=1}^{\infty }\nu (S_{j}),\end{aligned}}}$

as required.

The following property is a direct consequence of the definition of measure.

Lemma 2. Let ${\displaystyle \mu }$ be a measure, and ${\displaystyle S=\cup _{i=1}^{\infty }S_{i}}$, where

${\displaystyle S_{1}\subseteq \ldots \subseteq S_{i}\subseteq S_{i+1}\subseteq \ldots \subseteq S}$

is a non-decreasing chain with all its sets ${\displaystyle \mu }$-measurable. Then

${\displaystyle \mu (S)=\lim _{i}\mu (S_{i})}$.

Step 1. ${\displaystyle g_{n}=g_{n}(x)}$ is ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable, for every ${\displaystyle n\geq 1}$.

Indeed, since the Borel ${\displaystyle \sigma }$-algebra on ${\displaystyle \mathbb {R} \cup \{\pm \infty \}}$ is generated by the closed intervals ${\displaystyle \{[t,+\infty ]\}_{-\infty \leq t\leq +\infty }}$, it suffices to show that, ${\displaystyle g_{n}^{-1}([t,+\infty ])\in {\mathcal {F}}}$, for every ${\displaystyle t\in [-\infty ,+\infty ]}$, where ${\displaystyle g_{n}^{-1}([t,+\infty ])}$ denotes the inverse image of ${\displaystyle [t,+\infty ]}$ under ${\displaystyle g_{n}}$.

Observe that

${\displaystyle g_{n}(x)\geq t\quad \Leftrightarrow \quad {\Bigl (}\forall k\geq n\quad f_{k}(x)\geq t{\Bigr )}}$,

or equivalently,

${\displaystyle {\begin{aligned}g_{n}^{-1}([t,+\infty ])&=\left\{x\in X\mid g_{n}(x)\geq t\right\}\\[3pt]&=\bigcap _{k}\left\{x\in X\mid f_{k}(x)\geq t\right\}\\[3pt]&=\bigcap _{k}f_{k}^{-1}([t,+\infty ])\end{aligned}}}$

Note that every set on the right-hand side is from ${\displaystyle {\mathcal {F}}}$. Since, by definition, ${\displaystyle {\mathcal {F}}}$ is closed under countable intersections, we conclude that the left-hand side is also a member of ${\displaystyle {\mathcal {F}}}$. The ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurability of ${\displaystyle g_{n}}$ follows.

Step 2. Now, we want to show that the function ${\displaystyle f}$ is ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable.

If we were to use the monotone convergence theorem, the measurability of ${\displaystyle f}$ would follow easily from Remark 3.

Alternatively, using the technique from Step 1, it is enough to verify that ${\displaystyle f^{-1}([0,t])\in {\mathcal {F}}}$, for every ${\displaystyle t\in [-\infty ,+\infty ]}$. Since the sequence ${\displaystyle \{g_{n}(x)\}}$ pointwise non-decreases (see Remark 3), arguing as above, we get

${\displaystyle 0\leq f(x)\leq t\quad \Leftrightarrow \quad {\Bigl (}\forall n\quad 0\leq g_{n}(x)\leq t{\Bigr )}}$.

Due to the measurability of ${\displaystyle g_{n}}$, the above equivalency implies that

${\displaystyle f^{-1}([0,t])=\bigcap _{n}g_{n}^{-1}([0,t])\in {\mathcal {F}}}$.

End of Step 2.

The proof can proceed in two ways.

Proof using the monotone convergence theorem. By definition, ${\displaystyle \textstyle g_{n}(x)=\inf _{k\geq n}f_{k}(x)}$, so we have ${\displaystyle g_{n}\leq f_{n}}$, ${\displaystyle \forall n\in \mathbb {N} }$, and furthermore the sequence ${\displaystyle \{g_{n}(x)\}_{n}}$is non-decreasing ${\displaystyle \forall x\in X}$. Recall that ${\displaystyle f(x)=\liminf _{n\to \infty }f_{n}(x)=\lim _{n\to \infty }\inf _{k\geq n}f_{k}(x)}$, and therefore:

${\displaystyle {\begin{aligned}\int _{X}f\,d\mu &=\int _{X}\lim _{n}g_{n}\,d\mu \\&=\lim _{n}\int _{X}g_{n}\,d\mu \\&=\liminf _{n}\int _{X}g_{n}\,d\mu \\&\leq \liminf _{n}\int _{X}f_{n}\,d\mu ,\end{aligned}}}$

as required.

Independent proof. To prove the inequality without using the monotone convergence theorem, we need some extra machinery. Denote ${\displaystyle \operatorname {SF} (f)}$ the set of simple ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable functions ${\displaystyle s:X\to [0,\infty )}$ such that ${\displaystyle 0\leq s\leq f}$ on ${\displaystyle X}$.

Step 3. Given a simple function ${\displaystyle s\in \operatorname {SF} (f)}$ and a real number ${\displaystyle t\in (0,1)}$, define

${\displaystyle B_{k}^{s,t}=\{x\in X\mid t\cdot s(x)\leq g_{k}(x)\}\subseteq X.}$

Then ${\displaystyle B_{k}^{s,t}\in {\mathcal {F}}}$, ${\displaystyle B_{k}^{s,t}\subseteq B_{k+1}^{s,t}}$, and ${\displaystyle \textstyle X=\bigcup _{k}B_{k}^{s,t}}$.

Step 3a. To prove the first claim, let

${\displaystyle s=\sum _{i=1}^{m}c_{i}\cdot \mathbf {1} _{A_{i}},}$

for some finite collection of pairwise disjoint measurable sets ${\displaystyle A_{1},\ldots ,A_{m}\in {\mathcal {F}}}$ such that ${\displaystyle \textstyle X=\cup _{i=1}^{m}A_{i}}$, some (finite) real values ${\displaystyle c_{1},\ldots ,c_{m}}$, and ${\displaystyle \mathbf {1} _{A_{i}}}$ denoting the indicator function of the set ${\displaystyle A_{i}}$. Then

${\displaystyle B_{k}^{s,t}=\bigcup _{i=1}^{m}{\Bigl (}g_{k}^{-1}{\Bigl (}[t\cdot c_{i},+\infty ]{\Bigr )}\cap A_{i}{\Bigr )}}$.

Since the pre-image ${\displaystyle g_{k}^{-1}{\Bigl (}[t\cdot c_{i},+\infty ]{\Bigr )}}$ of the Borel set ${\displaystyle [t\cdot c_{i},+\infty ]}$ under the measurable function ${\displaystyle g_{k}}$ is measurable, and ${\displaystyle \sigma }$-algebras, by definition, are closed under finite intersection and unions, the first claim follows.

Step 3b. To prove the second claim, note that, for each ${\displaystyle k}$ and every ${\displaystyle x\in X}$, ${\displaystyle g_{k}(x)\leq g_{k+1}(x).}$

Step 3c. To prove the third claim, we show that ${\displaystyle \textstyle X\subseteq \bigcup _{k}B_{k}^{s,t}}$.

Indeed, if, to the contrary, ${\displaystyle \textstyle X\not \subseteq \bigcup _{k}B_{k}^{s,t}}$, then an element

${\displaystyle x_{0}\in X\setminus \bigcup _{k}B_{k}^{s,t}=\bigcap _{k}(X\setminus B_{k}^{s,t})}$

exists such that ${\displaystyle g_{k}(x_{0})<t\cdot s(x_{0})}$, for every ${\displaystyle k}$. Taking the limit as ${\displaystyle k\to \infty }$, get

${\displaystyle f(x_{0})\leq t\cdot s(x_{0})<s(x_{0}).}$

But by initial assumption, ${\displaystyle s\leq f}$. This is a contradiction.

Step 4. For every simple ${\displaystyle ({\mathcal {F}},\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable non-negative function ${\displaystyle s_{2}}$,

${\displaystyle \lim _{n}\int _{B_{n}^{s,t}}s_{2}\,d\mu =\int _{X}s_{2}\,d\mu .}$

To prove this, define ${\displaystyle \textstyle \nu (S)=\int _{S}s_{2}\,d\mu }$. By Lemma 1, ${\displaystyle \nu (S)}$ is a measure on ${\displaystyle \Omega }$. By "continuity from below" (Lemma 2),

${\displaystyle \lim _{n}\int _{B_{n}^{s,t}}s_{2}\,d\mu =\lim _{n}\nu (B_{n}^{s,t})=\nu (X)=\int _{X}s_{2}\,d\mu }$,

as required.

Step 5. We now prove that, for every ${\displaystyle s\in \operatorname {SF} (f)}$,

${\displaystyle \int _{X}s\,d\mu \leq \lim _{k}\int _{X}g_{k}\,d\mu }$.

Indeed, using the definition of ${\displaystyle B_{k}^{s,t}}$, the non-negativity of ${\displaystyle g_{k}}$, and the monotonicity of Lebesgue integral, we have

${\displaystyle \forall k\geq 1\qquad \int _{B_{k}^{s,t}}t\cdot s\,d\mu \leq \int _{B_{k}^{s,t}}g_{k}\,d\mu \leq \int _{X}g_{k}\,d\mu }$.

In accordance with Step 4, as ${\displaystyle k\to \infty }$ the inequality becomes

${\displaystyle t\int _{X}s\,d\mu \leq \lim _{k}\int _{X}g_{k}\,d\mu }$.

Taking the limit as ${\displaystyle t\uparrow 1}$ yields

${\displaystyle \int _{X}s\,d\mu \leq \lim _{k}\int _{X}g_{k}\,d\mu }$,

as required.

Step 6. To complete the proof, we apply the definition of Lebesgue integral to the inequality established in Step 5 and take into account that ${\displaystyle g_{n}\leq f_{n}}$:

${\displaystyle {\begin{aligned}\int _{X}f\,d\mu &=\sup _{s\in \operatorname {SF} (f)}\int _{X}s\,d\mu \\&\leq \lim _{k}\int _{X}g_{k}\,d\mu \\&=\liminf _{k}\int _{X}g_{k}\,d\mu \\&\leq \liminf _{k}\int _{X}f_{k}\,d\mu \end{aligned}}}$

The proof is complete.

We apply linearity of Lebesgue integral and Fatou's lemma to the sequence ${\displaystyle g-f_{n}.}$ Since ${\displaystyle \textstyle \int _{S}g\,d\mu <+\infty ,}$ this sequence is defined ${\displaystyle \mu }$-almost everywhere and non-negative.

Let f₁, f₂, . . . be a sequence of extended real-valued measurable functions defined on a measure space (S,Σ,μ). If there exists an integrable function g on S such that f_n ≥ −g for all n, then

${\displaystyle \int _{S}\liminf _{n\to \infty }f_{n}\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu .}$

Apply Fatou's lemma to the non-negative sequence given by f_n + g.

If in the previous setting the sequence f₁, f₂, . . . converges pointwise to a function f μ-almost everywhere on S, then

${\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu \,.}$

Note that f has to agree with the limit inferior of the functions f_n almost everywhere, and that the values of the integrand on a set of measure zero have no influence on the value of the integral.

The last assertion also holds, if the sequence f₁, f₂, . . . converges in measure to a function f.

There exists a subsequence such that

${\displaystyle \lim _{k\to \infty }\int _{S}f_{n_{k}}\,d\mu =\liminf _{n\to \infty }\int _{S}f_{n}\,d\mu .}$

Since this subsequence also converges in measure to f, there exists a further subsequence, which converges pointwise to f almost everywhere, hence the previous variation of Fatou's lemma is applicable to this subsubsequence.

In all of the above statements of Fatou's Lemma, the integration was carried out with respect to a single fixed measure μ. Suppose that μ_n is a sequence of measures on the measurable space (S,Σ) such that (see Convergence of measures)

${\displaystyle \mu _{n}(E)\to \mu (E),~\forall E\in {\mathcal {F}}.}$

Then, with f_n non-negative integrable functions and f being their pointwise limit inferior, we have

${\displaystyle \int _{S}f\,d\mu \leq \liminf _{n\to \infty }\int _{S}f_{n}\,d\mu _{n}.}$

Proof

We will prove something a bit stronger here. Namely, we will allow fn to converge μ-almost everywhere on a subset E of S. We seek to show that ${\displaystyle \int _{E}f\,d\mu \leq \liminf _{n\to \infty }\int _{E}f_{n}\,d\mu _{n}\,.}$ Let ${\displaystyle K=\{x\in E|f_{n}(x)\rightarrow f(x)\}}$. Then μ(E-K)=0 and ${\displaystyle \int _{E}f\,d\mu =\int _{E-K}f\,d\mu ,~~~\int _{E}f_{n}\,d\mu =\int _{E-K}f_{n}\,d\mu ~\forall n\in \mathbb {N} .}$ Thus, replacing E by E-K we may assume that fn converge to f pointwise on E. Next, note that for any simple function φ we have ${\displaystyle \int _{E}\phi \,d\mu =\lim _{n\to \infty }\int _{E}\phi \,d\mu _{n}.}$ Hence, by the definition of the Lebesgue Integral, it is enough to show that if φ is any non-negative simple function less than or equal to f, then ${\displaystyle \int _{E}\phi \,d\mu \leq \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu _{n}}$ Let a be the minimum non-negative value of φ. Define ${\displaystyle A=\{x\in E|\phi (x)>a\}}$ We first consider the case when ${\displaystyle \int _{E}\phi \,d\mu =\infty }$. We must have that μ(A) is infinite since ${\displaystyle \int _{E}\phi \,d\mu \leq M\mu (A),}$ where M is the (necessarily finite) maximum value of that φ attains. Next, we define ${\displaystyle A_{n}=\{x\in E|f_{k}(x)>a~\forall k\geq n\}.}$ We have that ${\displaystyle A\subseteq \bigcup _{n}A_{n}\Rightarrow \mu (\bigcup _{n}A_{n})=\infty .}$ But An is a nested increasing sequence of functions and hence, by the continuity from below μ, ${\displaystyle \lim _{n\rightarrow \infty }\mu (A_{n})=\infty .}$. Thus, ${\displaystyle \lim _{n\to \infty }\mu _{n}(A_{n})=\mu (A_{n})=\infty .}$. At the same time, ${\displaystyle \int _{E}f_{n}\,d\mu _{n}\geq a\mu _{n}(A_{n})\Rightarrow \liminf _{n\to \infty }\int _{E}f_{n}\,d\mu _{n}=\infty =\int _{E}\phi \,d\mu ,}$ proving the claim in this case. The remaining case is when ${\displaystyle \int _{E}\phi \,d\mu <\infty }$. We must have that μ(A) is finite. Denote, as above, by M the maximum value of φ and fix ε>0. Define ${\displaystyle A_{n}=\{x\in E|f_{k}(x)>(1-\epsilon )\phi (x)~\forall k\geq n\}.}$ Then An is a nested increasing sequence of sets whose union contains A. Thus, A-An is a decreasing sequence of sets with empty intersection. Since A has finite measure (this is why we needed to consider the two separate cases), ${\displaystyle \lim _{n\rightarrow \infty }\mu (A-A_{n})=0.}$ Thus, there exists n such that ${\displaystyle \mu (A-A_{k})<\epsilon ,~\forall k\geq n.}$ Therefore, since ${\displaystyle \lim _{n\to \infty }\mu _{n}(A-A_{k})=\mu (A-A_{k}),}$ there exists N such that ${\displaystyle \mu _{k}(A-A_{k})<\epsilon ,~\forall k\geq N.}$ Hence, for ${\displaystyle k\geq N}$ ${\displaystyle \int _{E}f_{k}\,d\mu _{k}\geq \int _{A_{k}}f_{k}\,d\mu _{k}\geq (1-\epsilon )\int _{A_{k}}\phi \,d\mu _{k}.}$ At the same time, ${\displaystyle \int _{E}\phi \,d\mu _{k}=\int _{A}\phi \,d\mu _{k}=\int _{A_{k}}\phi \,d\mu _{k}+\int _{A-A_{k}}\phi \,d\mu _{k}.}$ Hence, ${\displaystyle (1-\epsilon )\int _{A_{k}}\phi \,d\mu _{k}\geq (1-\epsilon )\int _{E}\phi \,d\mu _{k}-\int _{A-A_{k}}\phi \,d\mu _{k}.}$ Combining these inequalities gives that ${\displaystyle \int _{E}f_{k}\,d\mu _{k}\geq (1-\epsilon )\int _{E}\phi \,d\mu _{k}-\int _{A-A_{k}}\phi \,d\mu _{k}\geq \int _{E}\phi \,d\mu _{k}-\epsilon \left(\int _{E}\phi \,d\mu _{k}+M\right).}$ Hence, sending ε to 0 and taking the liminf in n, we get that ${\displaystyle \liminf _{n\rightarrow \infty }\int _{E}f_{n}\,d\mu _{k}\geq \int _{E}\phi \,d\mu ,}$ completing the proof.

Let X₁, X₂, . . . be a sequence of non-negative random variables on a probability space ${\displaystyle \scriptstyle (\Omega ,{\mathcal {F}},\mathbb {P} )}$ and let ${\displaystyle \scriptstyle {\mathcal {G}}\,\subset \,{\mathcal {F}}}$ be a sub-σ-algebra. Then

${\displaystyle \mathbb {E} {\Bigl [}\liminf _{n\to \infty }X_{n}\,{\Big |}\,{\mathcal {G}}{\Bigr ]}\leq \liminf _{n\to \infty }\,\mathbb {E} [X_{n}|{\mathcal {G}}]}$   almost surely.

Note: Conditional expectation for non-negative random variables is always well defined, finite expectation is not needed.

Besides a change of notation, the proof is very similar to the one for the standard version of Fatou's lemma above, however the monotone convergence theorem for conditional expectations has to be applied.

Let X denote the limit inferior of the X_n. For every natural number k define pointwise the random variable

${\displaystyle Y_{k}=\inf _{n\geq k}X_{n}.}$

Then the sequence Y₁, Y₂, . . . is increasing and converges pointwise to X. For k ≤ n, we have Y_k ≤ X_n, so that

${\displaystyle \mathbb {E} [Y_{k}|{\mathcal {G}}]\leq \mathbb {E} [X_{n}|{\mathcal {G}}]}$   almost surely

by the monotonicity of conditional expectation, hence

${\displaystyle \mathbb {E} [Y_{k}|{\mathcal {G}}]\leq \inf _{n\geq k}\mathbb {E} [X_{n}|{\mathcal {G}}]}$   almost surely,

because the countable union of the exceptional sets of probability zero is again a null set. Using the definition of X, its representation as pointwise limit of the Y_k, the monotone convergence theorem for conditional expectations, the last inequality, and the definition of the limit inferior, it follows that almost surely

${\displaystyle {\begin{aligned}\mathbb {E} {\Bigl [}\liminf _{n\to \infty }X_{n}\,{\Big |}\,{\mathcal {G}}{\Bigr ]}&=\mathbb {E} [X|{\mathcal {G}}]=\mathbb {E} {\Bigl [}\lim _{k\to \infty }Y_{k}\,{\Big |}\,{\mathcal {G}}{\Bigr ]}=\lim _{k\to \infty }\mathbb {E} [Y_{k}|{\mathcal {G}}]\\&\leq \lim _{k\to \infty }\inf _{n\geq k}\mathbb {E} [X_{n}|{\mathcal {G}}]=\liminf _{n\to \infty }\,\mathbb {E} [X_{n}|{\mathcal {G}}].\end{aligned}}}$

Let X₁, X₂, . . . be a sequence of random variables on a probability space ${\displaystyle \scriptstyle (\Omega ,{\mathcal {F}},\mathbb {P} )}$ and let ${\displaystyle \scriptstyle {\mathcal {G}}\,\subset \,{\mathcal {F}}}$ be a sub-σ-algebra. If the negative parts

${\displaystyle X_{n}^{-}:=\max\{-X_{n},0\},\qquad n\in {\mathbb {N} },}$

are uniformly integrable with respect to the conditional expectation, in the sense that, for ε > 0 there exists a c > 0 such that

${\displaystyle \mathbb {E} {\bigl [}X_{n}^{-}1_{\{X_{n}^{-}>c\}}\,|\,{\mathcal {G}}{\bigr ]}<\varepsilon ,\qquad {\text{for all }}n\in \mathbb {N} ,\,{\text{almost surely}}}$,

then

${\displaystyle \mathbb {E} {\Bigl [}\liminf _{n\to \infty }X_{n}\,{\Big |}\,{\mathcal {G}}{\Bigr ]}\leq \liminf _{n\to \infty }\,\mathbb {E} [X_{n}|{\mathcal {G}}]}$   almost surely.

Note: On the set where

${\displaystyle X:=\liminf _{n\to \infty }X_{n}}$

satisfies

${\displaystyle \mathbb {E} [\max\{X,0\}\,|\,{\mathcal {G}}]=\infty ,}$

the left-hand side of the inequality is considered to be plus infinity. The conditional expectation of the limit inferior might not be well defined on this set, because the conditional expectation of the negative part might also be plus infinity.

Let ε > 0. Due to uniform integrability with respect to the conditional expectation, there exists a c > 0 such that

${\displaystyle \mathbb {E} {\bigl [}X_{n}^{-}1_{\{X_{n}^{-}>c\}}\,|\,{\mathcal {G}}{\bigr ]}<\varepsilon \qquad {\text{for all }}n\in \mathbb {N} ,\,{\text{almost surely}}.}$

Since

${\displaystyle X+c\leq \liminf _{n\to \infty }(X_{n}+c)^{+},}$

where x⁺ := max{x,0} denotes the positive part of a real x, monotonicity of conditional expectation (or the above convention) and the standard version of Fatou's lemma for conditional expectations imply

${\displaystyle \mathbb {E} [X\,|\,{\mathcal {G}}]+c\leq \mathbb {E} {\Bigl [}\liminf _{n\to \infty }(X_{n}+c)^{+}\,{\Big |}\,{\mathcal {G}}{\Bigr ]}\leq \liminf _{n\to \infty }\mathbb {E} [(X_{n}+c)^{+}\,|\,{\mathcal {G}}]}$   almost surely.

Since

${\displaystyle (X_{n}+c)^{+}=(X_{n}+c)+(X_{n}+c)^{-}\leq X_{n}+c+X_{n}^{-}1_{\{X_{n}^{-}>c\}},}$

we have

${\displaystyle \mathbb {E} [(X_{n}+c)^{+}\,|\,{\mathcal {G}}]\leq \mathbb {E} [X_{n}\,|\,{\mathcal {G}}]+c+\varepsilon }$   almost surely,

hence

${\displaystyle \mathbb {E} [X\,|\,{\mathcal {G}}]\leq \liminf _{n\to \infty }\mathbb {E} [X_{n}\,|\,{\mathcal {G}}]+\varepsilon }$   almost surely.

This implies the assertion.