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Dominated convergence theorem
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<h2 id="statement">Statement</h2> <p>Lebesgue's dominated convergence theorem.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> Let ( f n ) {\displaystyle (f_{n})} be a sequence of <a href="/facts/Complex_number/2w7mqI7e">complex</a>-valued <a href="/facts/Measurable_function/NgIHnb1b">measurable functions</a> on a <a href="/facts/Measure_space/w9Dx4KY1">measure space</a> ( S , Σ , μ ) {\displaystyle (S,\Sigma ,\mu )} . Suppose that the sequence <a href="/facts/Pointwise_convergence/TXGpeyIK">converges pointwise</a> to a function f {\displaystyle f} i.e. </p> lim n → ∞ f n ( x ) = f ( x ) {\displaystyle \lim _{n\to \infty }f_{n}(x)=f(x)} <p>exists for every x ∈ S {\displaystyle x\in S} . Assume moreover that the sequence f n {\displaystyle f_{n}} is dominated by some integrable function g {\displaystyle g} in the sense that </p> | f n ( x ) | ≤ g ( x ) {\displaystyle |f_{n}(x)|\leq g(x)} <p>for all points x ∈ S {\displaystyle x\in S} and all n {\displaystyle n} in the index set. Then f n , f {\displaystyle f_{n},f} are integrable (in the <a href="/facts/Lebesgue_integration/f4AgvC6x">Lebesgue</a> sense) and </p> lim n → ∞ ∫ S f n d μ = ∫ S lim n → ∞ f n d μ = ∫ S f d μ {\displaystyle \lim _{n\to \infty }\int _{S}f_{n}\,d\mu =\int _{S}\lim _{n\to \infty }f_{n}d\mu =\int _{S}f\,d\mu } . <p>In fact, we have the stronger statement </p> lim n → ∞ ∫ S | f n − f | d μ = 0. {\displaystyle \lim _{n\to \infty }\int _{S}|f_{n}-f|\,d\mu =0.} <p> Remark 1. The statement " g {\displaystyle g} is integrable" means that the measurable function g {\displaystyle g} is Lebesgue integrable; i.e since g ≥ 0 {\displaystyle g\geq 0} . </p> ∫ S g d μ < ∞ . {\displaystyle \int _{S}g\,d\mu <\infty .} <p>Remark 2. The convergence of the sequence and domination by g {\displaystyle g} can be relaxed to hold only μ {\displaystyle \mu } -<a href="/facts/Almost_everywhere/1quRFtJE">almost everywhere</a> i.e. except possibly on a measurable set Z {\displaystyle Z} of μ {\displaystyle \mu } -measure 0 {\displaystyle 0} . In fact we can modify the functions f n {\displaystyle f_{n}} (hence its point wise limit f {\displaystyle f} ) to be 0 on Z {\displaystyle Z} without changing the value of the integrals. (If we insist on e.g. defining f {\displaystyle f} as the limit whenever it exists, we may end up with a <a href="/facts/Non-measurable_set/LhPvzlXT">non-measurable subset</a> within Z {\displaystyle Z} where convergence is violated if the measure space is <a href="/facts/Complete_measure/pB3gYaox">non complete</a>, and so f {\displaystyle f} might not be measurable. However, there is no harm in ignoring the limit inside the null set Z {\displaystyle Z} ). We can thus consider the f n {\displaystyle f_{n}} and f {\displaystyle f} as being defined except for a set of μ {\displaystyle \mu } -measure 0. </p><p>Remark 3. If μ ( S ) < ∞ {\displaystyle \mu (S)<\infty } , the condition that there is a dominating integrable function g {\displaystyle g} can be relaxed to <a href="/facts/Uniformly_integrable/0t0TR7pX">uniform integrability</a> of the sequence (<i>fn</i>), see <a href="/facts/Vitali_convergence_theorem/lpJ8PnpI">Vitali convergence theorem</a>. </p><p>Remark 4. While f {\displaystyle f} is Lebesgue integrable, it is not in general <a href="/facts/Riemann_integrable/vWGrPhVh">Riemann integrable</a>. For example, order the rationals in [ 0 , 1 ] {\displaystyle [0,1]} , and let f n {\displaystyle f_{n}} be defined on [ 0 , 1 ] {\displaystyle [0,1]} to take the value 1 on the first n rationals and 0 otherwise. Then f {\displaystyle f} is the <a href="/facts/Dirichlet_function/d9o708HR">Dirichlet function</a> on [ 0 , 1 ] {\displaystyle [0,1]} , which is not Riemann integrable but is Lebesgue integrable. </p><p> Remark 5 The stronger version of the dominated convergence theorem can be reformulated as: if a sequence of measurable complex functions f n {\displaystyle f_{n}} is almost everywhere pointwise convergent to a function f {\displaystyle f} and almost everywhere bounded in absolute value by an integrable function then f n → f {\displaystyle f_{n}\to f} in the <a href="/facts/Banach_space/sDEIk7EC">Banach space</a> L 1 ( S , μ ) {\displaystyle L_{1}(S,\mu )} </p> <h2 id="proof">Proof</h2> <p><a href="/facts/Without_loss_of_generality/gVP4dmEe">Without loss of generality</a>, one can assume that <i>f</i> is real, because one can split <i>f</i> into its real and imaginary parts (remember that a sequence of complex numbers converges <a href="/facts/If_and_only_if/bYSxGJ66">if and only if</a> both its real and imaginary counterparts converge) and apply the <a href="/facts/Triangle_inequality/KubLjkwr">triangle inequality</a> at the end. </p><p>Lebesgue's dominated convergence theorem is a special case of the <a href="/facts/Fatou%E2%80%93Lebesgue_theorem/62wQX6G2">Fatou–Lebesgue theorem</a>. Below, however, is a direct proof that uses <a href="/facts/Fatou%E2%80%99s_lemma/LwERbS3L">Fatou’s lemma</a> as the essential tool. </p><p>Since <i>f</i> is the pointwise limit of the sequence (<i>f</i><i>n</i>) of measurable functions that are dominated by <i>g</i>, it is also measurable and dominated by <i>g</i>, hence it is integrable. Furthermore, (these will be needed later), </p> | f − f n | ≤ | f | + | f n | ≤ 2 g {\displaystyle |f-f_{n}|\leq |f|+|f_{n}|\leq 2g} <p>for all <i>n</i> and </p> lim sup n → ∞ | f − f n | = 0. {\displaystyle \limsup _{n\to \infty }|f-f_{n}|=0.} <p>The second of these is trivially true (by the very definition of <i>f</i>). Using <a href="/facts/Lebesgue_integral/f4AgvC6x">linearity and monotonicity of the Lebesgue integral</a>, </p> | ∫ S f d μ − ∫ S f n d μ | = | ∫ S ( f − f n ) d μ | ≤ ∫ S | f − f n | d μ . {\displaystyle \left|\int _{S}{f\,d\mu }-\int _{S}{f_{n}\,d\mu }\right|=\left|\int _{S}{(f-f_{n})\,d\mu }\right|\leq \int _{S}{|f-f_{n}|\,d\mu }.} <p>By the <a href="/facts/Reverse_Fatou_lemma/LwERbS3L">reverse Fatou lemma</a> (it is here that we use the fact that |<i>f</i>−<i>fn</i>| is bounded above by an integrable function) </p> lim sup n → ∞ ∫ S | f − f n | d μ ≤ ∫ S lim sup n → ∞ | f − f n | d μ = 0 , {\displaystyle \limsup _{n\to \infty }\int _{S}|f-f_{n}|\,d\mu \leq \int _{S}\limsup _{n\to \infty }|f-f_{n}|\,d\mu =0,} <p>which implies that the limit exists and vanishes i.e. </p> lim n → ∞ ∫ S | f − f n | d μ = 0. {\displaystyle \lim _{n\to \infty }\int _{S}|f-f_{n}|\,d\mu =0.} <p>Finally, since </p> lim n → ∞ | ∫ S f d μ − ∫ S f n d μ | ≤ lim n → ∞ ∫ S | f − f n | d μ = 0. {\displaystyle \lim _{n\to \infty }\left|\int _{S}fd\mu -\int _{S}f_{n}d\mu \right|\leq \lim _{n\to \infty }\int _{S}|f-f_{n}|\,d\mu =0.} <p>we have that </p> lim n → ∞ ∫ S f n d μ = ∫ S f d μ . {\displaystyle \lim _{n\to \infty }\int _{S}f_{n}\,d\mu =\int _{S}f\,d\mu .} <p>The theorem now follows. </p><p>If the assumptions hold only μ-almost everywhere, then there exists a μ-null set <i>N</i> ∈ Σ such that the functions <i>fn</i> 1<i>S</i> \ <i>N</i> satisfy the assumptions everywhere on <i>S</i>. Then the function <i>f</i>(<i>x</i>) defined as the pointwise limit of <i>fn</i>(<i>x</i>) for <i>x</i> ∈ <i>S</i> \ <i>N</i> and by <i>f</i>(<i>x</i>) = 0 for <i>x</i> ∈ <i>N</i>, is measurable and is the pointwise limit of this modified function sequence. The values of these integrals are not influenced by these changes to the integrands on this μ-null set <i>N</i>, so the theorem continues to hold. </p><p>DCT holds even if <i>f</i><i>n</i> converges to <i>f</i> in measure (finite measure) and the dominating function is non-negative almost everywhere. </p> <h2 id="discussion-of-the-assumptions">Discussion of the assumptions</h2> <p>The assumption that the sequence is dominated by some integrable <i>g</i> cannot be dispensed with. This may be seen as follows: define <i>fn</i>(<i>x</i>) = <i>n</i> for <i>x</i> in the <a href="/facts/Interval_(mathematics)/e9se3vTe">interval</a> (0, 1/<i>n</i>] and <i>f</i><i>n</i>(<i>x</i>) = 0 otherwise. Any <i>g</i> which dominates the sequence must also dominate the pointwise <a href="/facts/Supremum/oXCKVopz">supremum</a> <i>h</i> = sup<i>n</i> <i>fn</i>. Observe that </p> ∫ 0 1 h ( x ) d x ≥ ∫ 1 m 1 h ( x ) d x = ∑ n = 1 m − 1 ∫ ( 1 n + 1 , 1 n ] h ( x ) d x ≥ ∑ n = 1 m − 1 ∫ ( 1 n + 1 , 1 n ] n d x = ∑ n = 1 m − 1 1 n + 1 → ∞ as m → ∞ {\displaystyle \int _{0}^{1}h(x)\,dx\geq \int _{\frac {1}{m}}^{1}{h(x)\,dx}=\sum _{n=1}^{m-1}\int _{\left({\frac {1}{n+1}},{\frac {1}{n}}\right]}{h(x)\,dx}\geq \sum _{n=1}^{m-1}\int _{\left({\frac {1}{n+1}},{\frac {1}{n}}\right]}{n\,dx}=\sum _{n=1}^{m-1}{\frac {1}{n+1}}\to \infty \qquad {\text{as }}m\to \infty } <p>by the divergence of the <a href="/facts/Harmonic_series_(mathematics)/8KGDHKTg">harmonic series</a>. Hence, the monotonicity of the Lebesgue integral tells us that there exists no integrable function which dominates the sequence on [0,1]. A direct calculation shows that integration and pointwise limit do not commute for this sequence: </p> ∫ 0 1 lim n → ∞ f n ( x ) d x = 0 ≠ 1 = lim n → ∞ ∫ 0 1 f n ( x ) d x , {\displaystyle \int _{0}^{1}\lim _{n\to \infty }f_{n}(x)\,dx=0\neq 1=\lim _{n\to \infty }\int _{0}^{1}f_{n}(x)\,dx,} <p>because the pointwise limit of the sequence is the <a href="/facts/Zero_function/4NBRtKHc">zero function</a>. Note that the sequence (<i>fn</i>) is not even <a href="/facts/Uniformly_integrable/0t0TR7pX">uniformly integrable</a>, hence also the <a href="/facts/Vitali_convergence_theorem/lpJ8PnpI">Vitali convergence theorem</a> is not applicable. </p> <h2 id="bounded-convergence-theorem">Bounded convergence theorem</h2> <p>One corollary to the dominated convergence theorem is the bounded convergence theorem, which states that if (<i>f</i><i>n</i>) is a sequence of <a href="/facts/Uniform_boundedness/nyxwk4Ix">uniformly bounded</a> <a href="/facts/Real_number/R02gw5Pb">complex</a>-valued <a href="/facts/Measurable_function/NgIHnb1b">measurable functions</a> which converges pointwise on a bounded <a href="/facts/Measure_space/w9Dx4KY1">measure space</a> (<i>S</i>, Σ, μ) (i.e. one in which μ(<i>S</i>) is finite) to a function <i>f</i>, then the limit <i>f</i> is an integrable function and </p> lim n → ∞ ∫ S f n d μ = ∫ S f d μ . {\displaystyle \lim _{n\to \infty }\int _{S}{f_{n}\,d\mu }=\int _{S}{f\,d\mu }.} <p>Remark: The pointwise convergence and uniform boundedness of the sequence can be relaxed to hold only μ-<a href="/facts/Almost_everywhere/1quRFtJE">almost everywhere</a>, provided the measure space (<i>S</i>, Σ, μ) is <a href="/facts/Measure_(mathematics)/yCq7zlI4">complete</a> or <i>f</i> is chosen as a measurable function which agrees μ-almost everywhere with the μ-almost everywhere existing pointwise limit. </p> <h3>Proof</h3> <p>Since the sequence is uniformly bounded, there is a real number <i>M</i> such that |<i>fn</i>(<i>x</i>)| ≤ <i>M</i> for all <i>x</i> ∈ <i>S</i> and for all <i>n</i>. Define <i>g</i>(<i>x</i>) = <i>M</i> for all <i>x</i> ∈ <i>S</i>. Then the sequence is dominated by <i>g</i>. Furthermore, <i>g</i> is integrable since it is a constant function on a set of finite measure. Therefore, the result follows from the dominated convergence theorem. </p><p>If the assumptions hold only μ-almost everywhere, then there exists a μ-null set <i>N</i> ∈ Σ such that the functions <i>fn</i>1<i>S</i>\<i>N</i> satisfy the assumptions everywhere on <i>S</i>. </p> <h2 id="dominated-convergence-in-lp-spaces-corollary">Dominated convergence in <i>L</i><i>p</i>-spaces (corollary)</h2> <p>Let ( Ω , A , μ ) {\displaystyle (\Omega ,{\mathcal {A}},\mu )} be a <a href="/facts/Measure_space/w9Dx4KY1">measure space</a>, 1 ≤ p < ∞ {\displaystyle 1\leq p<\infty } a real number and ( f n ) {\displaystyle (f_{n})} a sequence of A {\displaystyle {\mathcal {A}}} -measurable functions f n : Ω → C ∪ { ∞ } {\displaystyle f_{n}:\Omega \to \mathbb {C} \cup \{\infty \}} . </p><p>Assume the sequence ( f n ) {\displaystyle (f_{n})} converges μ {\displaystyle \mu } -almost everywhere to an A {\displaystyle {\mathcal {A}}} -measurable function f {\displaystyle f} , and is dominated by a g ∈ L p {\displaystyle g\in L^{p}} (cf. <a href="/facts/Lp_space/gbad0Rv1">Lp space</a>), i.e., for every natural number n {\displaystyle n} we have: | f n | ≤ g {\displaystyle |f_{n}|\leq g} , μ-almost everywhere. </p><p>Then all f n {\displaystyle f_{n}} as well as f {\displaystyle f} are in L p {\displaystyle L^{p}} and the sequence ( f n ) {\displaystyle (f_{n})} converges to f {\displaystyle f} in <a href="/facts/Lp-space/gbad0Rv1">the sense of L p {\displaystyle L^{p}} </a>, i.e.: </p> lim n → ∞ ‖ f n − f ‖ p = lim n → ∞ ( ∫ Ω | f n − f | p d μ ) 1 p = 0. {\displaystyle \lim _{n\to \infty }\|f_{n}-f\|_{p}=\lim _{n\to \infty }\left(\int _{\Omega }|f_{n}-f|^{p}\,d\mu \right)^{\frac {1}{p}}=0.} <p>Idea of the proof: Apply the original theorem to the function sequence h n = | f n − f | p {\displaystyle h_{n}=|f_{n}-f|^{p}} with the dominating function ( 2 g ) p {\displaystyle (2g)^{p}} . </p> <h2 id="extensions">Extensions</h2> <p>The dominated convergence theorem applies also to measurable functions with values in a <a href="/facts/Banach_space/sDEIk7EC">Banach space</a>, with the dominating function still being non-negative and integrable as above. The assumption of convergence almost everywhere can be weakened to require only <a href="/facts/Convergence_in_measure/SP7Hhyn3">convergence in measure</a>. </p><p>The dominated convergence theorem applies also to conditional expectations.<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Convergence_of_random_variables/pWdatFlY">Convergence of random variables</a>, <a href="/facts/Convergence_in_mean/pWdatFlY">Convergence in mean</a></li> <li><a href="/facts/Monotone_convergence_theorem/mYFH9yGY">Monotone convergence theorem</a> (does not require domination by an integrable function but assumes monotonicity of the sequence instead)</li> <li><a href="/facts/Scheff%C3%A9%27s_lemma/9GI8KKdL">Scheffé's lemma</a></li> <li><a href="/facts/Uniform_integrability/0t0TR7pX">Uniform integrability</a></li> <li><a href="/facts/Vitali_convergence_theorem/lpJ8PnpI">Vitali convergence theorem</a> (a generalization of Lebesgue's dominated convergence theorem)</li></ul> <h2 id="notes">Notes</h2> <ul><li>Bartle, R.G. (1995). <a href="https://books.google.com/books?id=8oNGAAAAYAAJ"><i>The Elements of Integration and Lebesgue Measure</i></a>. Wiley Interscience. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 9780471042228.</li> <li>Royden, H.L. (1988). <a href="https://books.google.com/books?id=J4k_AQAAIAAJ"><i>Real Analysis</i></a>. Prentice Hall. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 9780024041517.</li> <li>Weir, Alan J. (1973). "The Convergence Theorems". <i>Lebesgue Integration and Measure</i>. Cambridge: Cambridge University Press. pp. 93–118. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-521-08728-7.</li> <li><a href="/facts/David_Williams_(mathematician)/RCLoqkRp">Williams, D.</a> (1991). <i>Probability with martingales</i>. Cambridge University Press. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-521-40605-6.</li> <li>Zitkovic, Gordan (Fall 2013). <a href="https://web.ma.utexas.edu/users/gordanz/notes/conditional_expectation.pdf">"Lecture10: Conditional Expectation"</a> (PDF). Retrieved December 25, 2020.</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>For the real case, see Evans, Lawrence C; Gariepy, Ronald F (2015). Measure Theory and Fine Properties of Functions. CRC Press. pp. Theorem 1.19. <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Zitkovic 2013, Proposition 10.5. <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> </ol>