Throughout, N \mathbb{N} N denotes the natural numbers and R \mathbb{R} R the real numbers . We use the terms jointly Borel and the r r r -fold product Borel Ο \sigma Ο -algebra as defined in Joint Distribution, Expectations, and Block Independence for Independent Random Variables , and the proof is the replacement method of Lindeberg.
Step 1 (normalization). Set Y m = ( X m β ΞΌ ) / Ο Y_m=(X_m-\mu)/\sigma Y m β = ( X m β β ΞΌ ) / Ο and W n = ( Y 1 + β― + Y n ) / n W_n=(Y_1+\cdots+Y_n)/\sqrt{n} W n β = ( Y 1 β + β― + Y n β ) / n β , so that W n = ( S n β n ΞΌ ) / ( Ο n ) W_n=(S_n-n\mu)/(\sigma\sqrt{n}) W n β = ( S n β β n ΞΌ ) / ( Ο n β ) . The map h : R β R h:\mathbb{R}\to\mathbb{R} h : R β R , h ( x ) = ( x β ΞΌ ) / Ο h(x)=(x-\mu)/\sigma h ( x ) = ( x β ΞΌ ) / Ο , is continuous, hence Borel measurable by the generator criterion there, so each Y m = h ( X m ) Y_m=h(X_m) Y m β = h ( X m β ) is a random variable. The sequence ( Y m ) m β N (Y_m)_{m\in\mathbb{N}} ( Y m β ) m β N β is independent : for Borel sets B 1 , β¦ , B r B_1,\dots,B_r B 1 β , β¦ , B r β and indices m 1 < β― < m r m_1<\dots<m_r m 1 β < β― < m r β we have { Y m i β B i } = { X m i β h β 1 ( B i ) } \{Y_{m_i}\in B_i\}=\{X_{m_i}\in h^{-1}(B_i)\} { Y m i β β β B i β } = { X m i β β β h β 1 ( B i β )} , so the product identities required by that definition follow from independence of ( X m ) (X_m) ( X m β ) ; and the Y m Y_m Y m β are identically distributed, since P ( Y m β B ) = P ( X m β h β 1 ( B ) ) P(Y_m\in B)=P(X_m\in h^{-1}(B)) P ( Y m β β B ) = P ( X m β β h β 1 ( B )) depends only on the common distribution of the X m X_m X m β . By linearity (Linearity and Monotonicity of the Lebesgue Integral ) and the definition of variance , Y 1 Y_1 Y 1 β and Y 1 2 Y_1^{2} Y 1 2 β are integrable with
E [ Y 1 ] = E [ X 1 ] β ΞΌ Ο = 0 , E [ Y 1 2 ] = Var β‘ ( X 1 ) Ο 2 = 1. \mathbb{E}[Y_1]=\frac{\mathbb{E}[X_1]-\mu}{\sigma}=0,\qquad \mathbb{E}[Y_1^{2}]=\frac{\operatorname{Var}(X_1)}{\sigma^{2}}=1. E [ Y 1 β ] = Ο E [ X 1 β ] β ΞΌ β = 0 , E [ Y 1 2 β ] = Ο 2 Var ( X 1 β ) β = 1.
Let ΞΌ Y \mu_Y ΞΌ Y β denote the distribution of Y 1 Y_1 Y 1 β and N N N the standard normal distribution .
By claim 2 of Smooth Test Function Criterion for Convergence in Distribution it suffices to prove that E [ f ( W n ) ] β E [ f ( Z ) ] \mathbb{E}[f(W_n)]\to\mathbb{E}[f(Z)] E [ f ( W n β )] β E [ f ( Z )] for every admissible test function f f f in the sense of that theorem; once W n β Z W_n\to Z W n β β Z in distribution is established, the displayed convergence in the statement holds at every t β R t\in\mathbb{R} t β R because Ξ¦ \Phi Ξ¦ is continuous everywhere (claim 3 of The Gaussian Weight Defines a Probability Distribution ). Fix such an f f f and set M 2 = sup β‘ β£ f β² β² β£ M_2=\sup|f''| M 2 β = sup β£ f β²β² β£ and M 3 = sup β‘ β£ f β² β² β² β£ M_3=\sup|f'''| M 3 β = sup β£ f β²β²β² β£ , finite by admissibility; all expectations of compositions of f f f (or of the bounded functions f β² , f β² β² f',f'' f β² , f β²β² ) with random variables below are defined by claim 1 of Smooth Test Function Criterion for Convergence in Distribution and its argument.
Step 2 (auxiliary space and reduction). Apply Existence of Independent Sequences with Prescribed Distributions to the sequence of probability measures given by Ξ½ 2 k β 1 = ΞΌ Y \nu_{2k-1}=\mu_Y Ξ½ 2 k β 1 β = ΞΌ Y β and Ξ½ 2 k = N \nu_{2k}=N Ξ½ 2 k β = N for k β₯ 1 k\ge 1 k β₯ 1 : there are a probability space ( Ξ© β² , F β² , P β² ) (\Omega',\mathcal{F}',P') ( Ξ© β² , F β² , P β² ) and an independent sequence ( U m ) m β N (U_m)_{m\in\mathbb{N}} ( U m β ) m β N β on it with these distributions. Write Y k β² = U 2 k β 1 Y'_k=U_{2k-1} Y k β² β = U 2 k β 1 β and Z k = U 2 k Z_k=U_{2k} Z k β = U 2 k β ; every finite subfamily of an independent family is independent directly from Independence of Events and of Random Variables , so for each n n n the variables Y 1 β² , β¦ , Y n β² , Z 1 , β¦ , Z n Y'_1,\dots,Y'_n,Z_1,\dots,Z_n Y 1 β² β , β¦ , Y n β² β , Z 1 β , β¦ , Z n β are independent, each Y k β² Y'_k Y k β² β with distribution ΞΌ Y \mu_Y ΞΌ Y β and each Z k Z_k Z k β standard normal. Expectation on ( Ξ© β² , F β² , P β² ) (\Omega',\mathcal{F}',P') ( Ξ© β² , F β² , P β² ) is written E β² \mathbb{E}' E β² . Set
W n β² = Y 1 β² + β― + Y n β² n , G n = Z 1 + β― + Z n n . W'_n=\frac{Y'_1+\cdots+Y'_n}{\sqrt{n}},\qquad G_n=\frac{Z_1+\cdots+Z_n}{\sqrt{n}}. W n β² β = n β Y 1 β² β + β― + Y n β² β β , G n β = n β Z 1 β + β― + Z n β β .
By claim 4 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables , the map w n ( x 1 , β¦ , x n ) = f ( ( x 1 + β― + x n ) / n ) w_n(x_1,\dots,x_n)=f\bigl((x_1+\cdots+x_n)/\sqrt{n}\bigr) w n β ( x 1 β , β¦ , x n β ) = f ( ( x 1 β + β― + x n β ) / n β ) is jointly Borel (the addition map is jointly Borel, and t β¦ f ( t / n ) t\mapsto f(t/\sqrt{n}) t β¦ f ( t / n β ) is continuous, hence Borel), and it is bounded. By claim 2 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables , applied on ( Ξ© , F , P ) (\Omega,\mathcal{F},P) ( Ξ© , F , P ) to the independent tuple Y 1 , β¦ , Y n Y_1,\dots,Y_n Y 1 β , β¦ , Y n β and on ( Ξ© β² , F β² , P β² ) (\Omega',\mathcal{F}',P') ( Ξ© β² , F β² , P β² ) to Y 1 β² , β¦ , Y n β² Y'_1,\dots,Y'_n Y 1 β² β , β¦ , Y n β² β β all entries with distribution ΞΌ Y \mu_Y ΞΌ Y β β both E [ f ( W n ) ] \mathbb{E}[f(W_n)] E [ f ( W n β )] and E β² [ f ( W n β² ) ] \mathbb{E}'[f(W'_n)] E β² [ f ( W n β² β )] equal the integral of w n w_n w n β against the n n n -fold product of ΞΌ Y \mu_Y ΞΌ Y β with itself, so
E [ f ( W n ) ] = E β² [ f ( W n β² ) ] . \mathbb{E}[f(W_n)]=\mathbb{E}'[f(W'_n)]. E [ f ( W n β )] = E β² [ f ( W n β² β )] .
By claim 3 of Moments and Stability of the Standard Normal Distribution , G n G_n G n β is standard normal; since f f f is bounded and Borel, Change of Variables for Expectations applied to G n G_n G n β and to Z Z Z (which share the distribution N N N ) gives E β² [ f ( G n ) ] = E [ f ( Z ) ] \mathbb{E}'[f(G_n)]=\mathbb{E}[f(Z)] E β² [ f ( G n β )] = E [ f ( Z )] . It therefore suffices to show
E β² [ f ( W n β² ) ] β E β² [ f ( G n ) ] βΆ 0 ( n β β ) . \mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]\longrightarrow 0\qquad(n\to\infty). E β² [ f ( W n β² β )] β E β² [ f ( G n β )] βΆ 0 ( n β β ) .
Step 3 (two Taylor bounds). For x , u β R x,u\in\mathbb{R} x , u β R put
e ( x , u ) = f ( x + u ) β f ( x ) β f β² ( x ) β u β 1 2 f β² β² ( x ) β u 2 . e(x,u)=f(x+u)-f(x)-f'(x)\,u-\tfrac{1}{2}f''(x)\,u^{2}. e ( x , u ) = f ( x + u ) β f ( x ) β f β² ( x ) u β 2 1 β f β²β² ( x ) u 2 .
By Taylor Expansion with Third-Order Remainder Bound , β£ e ( x , u ) β£ β€ M 3 β£ u β£ 3 / 6 |e(x,u)|\le M_3|u|^{3}/6 β£ e ( x , u ) β£ β€ M 3 β β£ u β£ 3 /6 . Moreover, carrying out Steps 1 and 2 of the proof of that lemma one order lower β Fundamental Theorem of Calculus, Part II in One Dimension applied to f β² f' f β² gives β£ f β² ( x + t ) β f β² ( x ) β£ β€ M 2 β£ t β£ |f'(x+t)-f'(x)|\le M_2|t| β£ f β² ( x + t ) β f β² ( x ) β£ β€ M 2 β β£ t β£ , and then applied to f f f gives β£ f ( x + u ) β f ( x ) β f β² ( x ) u β£ β€ M 2 u 2 / 2 |f(x+u)-f(x)-f'(x)u|\le M_2u^{2}/2 β£ f ( x + u ) β f ( x ) β f β² ( x ) u β£ β€ M 2 β u 2 /2 (for negative increments exchange the endpoints, as there) β while β£ 1 2 f β² β² ( x ) u 2 β£ β€ M 2 u 2 / 2 |\tfrac12 f''(x)u^{2}|\le M_2u^{2}/2 β£ 2 1 β f β²β² ( x ) u 2 β£ β€ M 2 β u 2 /2 ; adding these two bounds,
β£ e ( x , u ) β£ Β β€ Β g ( u ) : = min β‘ { M 2 β u 2 , Β 1 6 M 3 β β£ u β£ 3 } . |e(x,u)|\ \le\ g(u):=\min\Bigl\{M_2\,u^{2},\ \tfrac{1}{6}M_3\,|u|^{3}\Bigr\}. β£ e ( x , u ) β£ Β β€ Β g ( u ) := min { M 2 β u 2 , Β 6 1 β M 3 β β£ u β£ 3 } .
The function g g g is continuous (a minimum of two continuous functions), hence Borel, and g β₯ 0 g\ge 0 g β₯ 0 .
Step 4 (the replacement estimate). Fix n β₯ 1 n\ge 1 n β₯ 1 and k β { 1 , β¦ , n } k\in\{1,\dots,n\} k β { 1 , β¦ , n } , and define on ( Ξ© β² , F β² , P β² ) (\Omega',\mathcal{F}',P') ( Ξ© β² , F β² , P β² )
R k = Z 1 + β― + Z k β 1 + Y k + 1 β² + β― + Y n β² n , H k = R k + Y k β² n , H ~ k = R k + Z k n . R_k=\frac{Z_1+\cdots+Z_{k-1}+Y'_{k+1}+\cdots+Y'_n}{\sqrt{n}},\qquad H_k=R_k+\frac{Y'_k}{\sqrt{n}},\qquad \widetilde{H}_k=R_k+\frac{Z_k}{\sqrt{n}}. R k β = n β Z 1 β + β― + Z k β 1 β + Y k + 1 β² β + β― + Y n β² β β , H k β = R k β + n β Y k β² β β , H k β = R k β + n β Z k β β .
Then H 1 = W n β² H_1=W'_n H 1 β = W n β² β , H ~ n = G n \widetilde{H}_n=G_n H n β = G n β , and H ~ k = H k + 1 \widetilde{H}_k=H_{k+1} H k β = H k + 1 β for 1 β€ k < n 1\le k<n 1 β€ k < n , so
E β² [ f ( W n β² ) ] β E β² [ f ( G n ) ] = β k = 1 n ( E β² [ f ( H k ) ] β E β² [ f ( H ~ k ) ] ) . \mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]=\sum_{k=1}^{n}\bigl(\mathbb{E}'[f(H_k)]-\mathbb{E}'[f(\widetilde{H}_k)]\bigr). E β² [ f ( W n β² β )] β E β² [ f ( G n β )] = k = 1 β n β ( E β² [ f ( H k β )] β E β² [ f ( H k β )] ) .
Taylor expansion at x = R k x=R_k x = R k β (Step 3) gives the pointwise identities
f ( H k ) = f ( R k ) + f β² ( R k ) Y k β² n + f β² β² ( R k ) 2 β
( Y k β² ) 2 n + e ( R k , Y k β² n ) , f(H_k)=f(R_k)+f'(R_k)\frac{Y'_k}{\sqrt{n}}+\frac{f''(R_k)}{2}\cdot\frac{(Y'_k)^{2}}{n}+e\Bigl(R_k,\frac{Y'_k}{\sqrt{n}}\Bigr), f ( H k β ) = f ( R k β ) + f β² ( R k β ) n β Y k β² β β + 2 f β²β² ( R k β ) β β
n ( Y k β² β ) 2 β + e ( R k β , n β Y k β² β β ) ,
and the same with Z k Z_k Z k β in place of Y k β² Y'_k Y k β² β for H ~ k \widetilde{H}_k H k β .
The variable R k R_k R k β is a jointly Borel function (claims 4 and hence 3 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables ) of the block of the independent family ( U m ) (U_m) ( U m β ) with indices in I k = { 2 i : i < k } βͺ { 2 j β 1 : j > k , Β j β€ n } I_k=\{2i:i<k\}\cup\{2j-1:j>k,\ j\le n\} I k β = { 2 i : i < k } βͺ { 2 j β 1 : j > k , Β j β€ n } , while Y k β² Y'_k Y k β² β and Z k Z_k Z k β are functions of the singleton blocks { 2 k β 1 } \{2k-1\} { 2 k β 1 } and { 2 k } \{2k\} { 2 k } , each disjoint from I k I_k I k β . Composing with the Borel maps f β² f' f β² , f β² β² f'' f β²β² , and t β¦ t 2 t\mapsto t^{2} t β¦ t 2 (claim 4), claim 3 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables shows that each of the pairs ( f β² ( R k ) , β Y k β² ) (f'(R_k),\,Y'_k) ( f β² ( R k β ) , Y k β² β ) , ( f β² β² ( R k ) , β ( Y k β² ) 2 ) (f''(R_k),\,(Y'_k)^{2}) ( f β²β² ( R k β ) , ( Y k β² β ) 2 ) , ( f β² ( R k ) , β Z k ) (f'(R_k),\,Z_k) ( f β² ( R k β ) , Z k β ) , ( f β² β² ( R k ) , β Z k 2 ) (f''(R_k),\,Z_k^{2}) ( f β²β² ( R k β ) , Z k 2 β ) consists of independent random variables. The first factors are bounded, hence integrable on a probability space, and the second factors are integrable: Y k β² Y'_k Y k β² β and ( Y k β² ) 2 (Y'_k)^{2} ( Y k β² β ) 2 share the integrability and moments of Y 1 Y_1 Y 1 β by Change of Variables for Expectations (equal distributions), and Z k Z_k Z k β , Z k 2 Z_k^{2} Z k 2 β are integrable with E β² [ Z k ] = 0 \mathbb{E}'[Z_k]=0 E β² [ Z k β ] = 0 , E β² [ Z k 2 ] = 1 \mathbb{E}'[Z_k^{2}]=1 E β² [ Z k 2 β ] = 1 by claim 1 of Moments and Stability of the Standard Normal Distribution . Hence Expectation of a Product of Independent Random Variables gives
E β² [ f β² ( R k ) β Y k β² ] = E β² [ f β² ( R k ) ] β
0 = E β² [ f β² ( R k ) β Z k ] , E β² [ f β² β² ( R k ) β ( Y k β² ) 2 ] = E β² [ f β² β² ( R k ) ] β
1 = E β² [ f β² β² ( R k ) β Z k 2 ] . \mathbb{E}'\bigl[f'(R_k)\,Y'_k\bigr]=\mathbb{E}'[f'(R_k)]\cdot 0=\mathbb{E}'\bigl[f'(R_k)\,Z_k\bigr],\qquad \mathbb{E}'\bigl[f''(R_k)\,(Y'_k)^{2}\bigr]=\mathbb{E}'[f''(R_k)]\cdot 1=\mathbb{E}'\bigl[f''(R_k)\,Z_k^{2}\bigr]. E β² [ f β² ( R k β ) Y k β² β ] = E β² [ f β² ( R k β )] β
0 = E β² [ f β² ( R k β ) Z k β ] , E β² [ f β²β² ( R k β ) ( Y k β² β ) 2 ] = E β² [ f β²β² ( R k β )] β
1 = E β² [ f β²β² ( R k β ) Z k 2 β ] .
Taking expectations in the two Taylor identities (each term integrable as just noted, the remainder terms being differences of integrable terms) and subtracting, the zeroth-, first-, and second-order terms cancel, so by linearity and monotonicity (Linearity and Monotonicity of the Lebesgue Integral ) together with the bound β£ e ( R k , u ) β£ β€ g ( u ) |e(R_k,u)|\le g(u) β£ e ( R k β , u ) β£ β€ g ( u ) of Step 3,
β£ E β² [ f ( H k ) ] β E β² [ f ( H ~ k ) ] β£ Β β€ Β E β² [ g ( Y k β² n ) ] + E β² [ g ( Z k n ) ] . \bigl|\mathbb{E}'[f(H_k)]-\mathbb{E}'[f(\widetilde{H}_k)]\bigr|\ \le\ \mathbb{E}'\Bigl[g\Bigl(\frac{Y'_k}{\sqrt{n}}\Bigr)\Bigr]+\mathbb{E}'\Bigl[g\Bigl(\frac{Z_k}{\sqrt{n}}\Bigr)\Bigr]. β E β² [ f ( H k β )] β E β² [ f ( H k β )] β Β β€ Β E β² [ g ( n β Y k β² β β ) ] + E β² [ g ( n β Z k β β ) ] .
For the Gaussian term, g ( u ) β€ M 3 β£ u β£ 3 / 6 g(u)\le M_3|u|^{3}/6 g ( u ) β€ M 3 β β£ u β£ 3 /6 and claim 1 of Moments and Stability of the Standard Normal Distribution give
E β² [ g ( Z k n ) ] Β β€ Β M 3 β Ξ³ 6 β n n , Ξ³ = E β² [ β£ Z 1 β£ 3 ] < β , \mathbb{E}'\Bigl[g\Bigl(\frac{Z_k}{\sqrt{n}}\Bigr)\Bigr]\ \le\ \frac{M_3\,\gamma}{6\,n\sqrt{n}},\qquad \gamma=\mathbb{E}'\bigl[|Z_1|^{3}\bigr]<\infty, E β² [ g ( n β Z k β β ) ] Β β€ Β 6 n n β M 3 β Ξ³ β , Ξ³ = E β² [ β£ Z 1 β β£ 3 ] < β ,
the value Ξ³ \gamma Ξ³ being shared by all the Z k Z_k Z k β (Change of Variables for Expectations , equal distributions). For the sample term, the Y k β² Y'_k Y k β² β share the distribution ΞΌ Y \mu_Y ΞΌ Y β and u β¦ g ( u / n ) u\mapsto g(u/\sqrt{n}) u β¦ g ( u / n β ) is nonnegative Borel, so E β² [ g ( Y k β² / n ) ] = E β² [ g ( Y 1 β² / n ) ] \mathbb{E}'[g(Y'_k/\sqrt{n})]=\mathbb{E}'[g(Y'_1/\sqrt{n})] E β² [ g ( Y k β² β / n β )] = E β² [ g ( Y 1 β² β / n β )] for every k k k (Change of Variables for Expectations ). Summing the n n n replacement estimates,
β£ E β² [ f ( W n β² ) ] β E β² [ f ( G n ) ] β£ Β β€ Β n β E β² [ g ( Y 1 β² n ) ] + M 3 β Ξ³ 6 n . \bigl|\mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]\bigr|\ \le\ n\,\mathbb{E}'\Bigl[g\Bigl(\frac{Y'_1}{\sqrt{n}}\Bigr)\Bigr]+\frac{M_3\,\gamma}{6\sqrt{n}}. β E β² [ f ( W n β² β )] β E β² [ f ( G n β )] β Β β€ Β n E β² [ g ( n β Y 1 β² β β ) ] + 6 n β M 3 β Ξ³ β .
Step 5 (vanishing of the error). For y β R y\in\mathbb{R} y β R ,
n β g ( y n ) = min β‘ { M 2 β y 2 , Β M 3 β β£ y β£ 3 6 n } = : g n ( y ) . n\,g\Bigl(\frac{y}{\sqrt{n}}\Bigr)=\min\Bigl\{M_2\,y^{2},\ \frac{M_3\,|y|^{3}}{6\sqrt{n}}\Bigr\}=:g_n(y). n g ( n β y β ) = min { M 2 β y 2 , Β 6 n β M 3 β β£ y β£ 3 β } =: g n β ( y ) .
Each g n g_n g n β is continuous, hence Borel; 0 β€ g n ( y ) β€ M 2 y 2 0\le g_n(y)\le M_2y^{2} 0 β€ g n β ( y ) β€ M 2 β y 2 ; and g n ( y ) β 0 g_n(y)\to 0 g n β ( y ) β 0 as n β β n\to\infty n β β for every fixed y y y , since the second entry of the minimum tends to 0 0 0 . Hence g n ( Y 1 β² ) β 0 g_n(Y'_1)\to 0 g n β ( Y 1 β² β ) β 0 pointwise on Ξ© β² \Omega' Ξ© β² , dominated by the integrable function M 2 β ( Y 1 β² ) 2 M_2\,(Y'_1)^{2} M 2 β ( Y 1 β² β ) 2 (Step 1 and equality of distributions). By Dominated Convergence Theorem ,
n β E β² [ g ( Y 1 β² n ) ] = E β² [ g n ( Y 1 β² ) ] βΆ 0 , n\,\mathbb{E}'\Bigl[g\Bigl(\frac{Y'_1}{\sqrt{n}}\Bigr)\Bigr]=\mathbb{E}'[g_n(Y'_1)]\longrightarrow 0, n E β² [ g ( n β Y 1 β² β β ) ] = E β² [ g n β ( Y 1 β² β )] βΆ 0 ,
and also M 3 Ξ³ / ( 6 n ) β 0 M_3\gamma/(6\sqrt{n})\to 0 M 3 β Ξ³ / ( 6 n β ) β 0 . Combining with Step 2,
E [ f ( W n ) ] β E [ f ( Z ) ] = E β² [ f ( W n β² ) ] β E β² [ f ( G n ) ] βΆ 0. \mathbb{E}[f(W_n)]-\mathbb{E}[f(Z)]=\mathbb{E}'[f(W'_n)]-\mathbb{E}'[f(G_n)]\longrightarrow 0. E [ f ( W n β )] β E [ f ( Z )] = E β² [ f ( W n β² β )] β E β² [ f ( G n β )] βΆ 0.
Since f f f was an arbitrary admissible test function, claim 2 of Smooth Test Function Criterion for Convergence in Distribution yields W n β Z W_n\to Z W n β β Z in distribution in the sense of Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution , and as noted in Step 1 the convergence P ( W n β€ t ) β Ξ¦ ( t ) P(W_n\le t)\to\Phi(t) P ( W n β β€ t ) β Ξ¦ ( t ) holds at every t β R t\in\mathbb{R} t β R because Ξ¦ \Phi Ξ¦ is continuous everywhere. β \blacksquare β